looking at this actually asserts my concern in my original comment because in fact cos(2x) = +/- sqrt(1-sin^2(2x)) which leads to multiple valid f(x) in the end (both A and B). In my original comment I even go over where it is possible for f(x) = 0 which both choice A and B miss since the domain of A and B is [-1,0) union (0,1]
@@florianbasier you are missing the fact that cos(2x) can be negative so = (1 - cos(2x))/(sin(2x)) = (1 - √(1 - sin²(2x)) / (sin(2x)) are not equivalent those two lines suggest cos(2x) = sqrt(1 - sin^2(x)) only and since sqrt() is non negative, it apparently implies cos(2x) is non negative, which it is not. (1 + sqrt(1-x^2))/x gives 1/tan(x) ON THE CONDITION that sqrt(1-x^2) = cos(2x), this only works when cos(2x) is non negative..., so when cos(2x) is negative, it's fair game and gives tan(x).
What if you don't have enough time for the rest of questions because you have to look deep enough in each option to make sure you are not missing something
This is correct in the range from [-pi/4, pi/4] (except zero, but the limit is correct). It is not correct in general for any real x. The reason this solution fails is that the image of 1/2 arcsin (x) only covers this range, so the substitution does not work in general.
I wish I would have had you as my math teacher way back when I was back in high school way back in the early eighties. I think I learned more from your videos than I ever did sitting in class and struggling through what was being taught in class
To claim that x is the half of arcsin of t, you need to know that x is in [-pi/4, pi/4] which is not in the question. Without that information, there may be other solutions using x=1/2sin^-1(t) + pi for example. It is especially problematic since the question is framed as an implication. Students have difficulties with inverse trigonometric functions and logic, so it is a disservice to them
This is an interesting problem, but it is not well defined. The answer is not necessarily B. It can also be A depending on what domain values we consider. Remember there is a square root involved. When you take square roots, there is ± depending on the domain/range. Since there is "None of these" in the answer, the correct choice would be D as there is no answer saying "Not enough information given". Thank you for bringing this problem.
Since f(sin(2x))=tanx, I opted to just evaluate each expression at x -> sin(2x) and see which answer simplified to tanx. No 1/2 angle or inverse identities to worry about that way; not that I don't know them, or couldn't figure them out. I thought it would be more straightforward in this case.
if we do the multiple choice method of just going through each option one by one then we can see A is valid on some condition (similarly you will find B is valid on some condition) for A: f(x) = (1 + sqrt(1-x^2))/x => f(sin(2x)) = (1 + sqrt(1 - sin^2(2x)))/sin(2x) => f(sin(2x)) = (1 + sqrt(cos^2(2x)))/sin(2x) => f(sin(2x)) = (1 + |cos(2x)|)/sin(2x) => f(sin(2x)) = (1 - cos(2x))/sin(2x) when cos(2x) f(sin(2x)) = tan(x) So A is valid provided cos(2x) = 0 Both of these miss f(0) = 0 but if u analyse the limits, B approaches 0 when x approaches 0.
by the very official Multiple Choice Theorem, it's b.) two answers with 1 at the start, two answers with x^1 at the bottom, and two answers subtracting the radical. the only answer containing all 3 is b. unless it's d.), in which case, that stinks.
We can cheat somewhat. consider x so small sin x ≈ x, above zero. Then a) = 1 + 1- / 0+ = 2- / 0+ = inf b) = 1 - 1- /0+ = 0-/0- (can't be thrown out, need investigate further) C) = 0+ - 1- / 0+ = -1±/0+ = inf. So it's b) or d) Also I'm on phone, so sorry for notation. 0+ mean as "0 + very small number" and / 0+ = inf means "we so close to zero.dividing gets really big number" - tan(0) is zero, not inf
We can write sin(2x) as 2tanx/ 1+ tan^2x and then take tanx = u so then f(2u/1+u^2) = u then u can find f(x) equals what by using 2u/1+u^2 = x and find u in terms of x so I got 2 values of f(x) but the problem is both are in the options so pls tell why the 1st option is wrong if we use this way
Well, due to 'just' looking at the situation in a right triangle, the angles for which a cosine is negative are avoided. But a negative square root? Let's look at complex numbers... Let's find a complex number z, such that: Arg(z^(1/2))=pi=Arg(z)/2 => Arg(z)=2pi, but an argument (aka fase) of a complex number needs to be an element of (-pi,pi] So let's multiply z with 1=e^(-i2pi) and make arg(z) equal to 0, and thus, z is just a positive real number. But Arg(z)=0 also means Arg(z)/2=0 which contradicts Arg(z)/2=pi, so a negative square root indeed doesn't exist. /s
@@TaladrisKpop BlackPenRedPen literally spoke about those right triangles in the video itself. But you're right that I should have named the triangles right triangles and shall correct my message above.
Sadly, if you actually check B, it only gives the right value over half the reals. A gives the right value over the other half. Since no answer gives the right value over all reals, the correct answer is D.
2:25 ive never encountered this type of question. can someone confirm this is correct: When you say f(sin(2x)) = tan x And you want to solve f(x) That means you know the output for the unknown function f(sin(2x)) with x substituted for sin(2x) is the same function as tan(x) So when you want the function f(x) you need to find a way to rewrite f(sin(2x)) as f(x) by defining sin(2x) as a single variable So sin(2x) = t But now you you need to define the other side, tan x to be equal. So you use sin(2x) = t and solve for x, so you can fill that in tan (x) and now you only have t’s since you defined al x’s in terms of t. And then you replace t with x with what matters is that there is 1 of the same variable on both sides, and the left side is f(x) Is this correct reasoning?
If we are talking about the domain (-pi/2, pi/2): Isn't option B true just for angles on the domain [-pi/4, 0) and (0, pi/4], And then option A is true for angles on the domain (-pi/2, -pi/4] and [pi/4, pi/2)?
You forgot that the input into that f(x) is restricted to sin(2x) in order for the function to equal tan(x). It breaks the function at the correct points to make tan x be correct. When sin(2x) = 0, cos(x) = 0. Your claim is that the resultant function needs to have that domain when it is the composition of the resultant function and sin(2x) does.
The answer depends on the x domain we are talking about for x € (-pi/4 +2kpi ,pi/4 +2kpi) U (3pi/4 +2kpi , 5pi/4 +2kpi) where k is any integer k...answer B holds.....but for x € ( pi/4 +2kpi , 3pi/4 +2kpi) U (5pi/4+ 2kpi, 7pi/4 +2kpi) where k is an integer...answer A holds....so for lim x -> pi/2 (-)(left handed limit) ....sin (2x) will approach zero but sill be larger than zero so giving us the correct result of tan x diverging +infinity.... (notice answer B won't work only A will work cause of the domain)
This is almost the Weierstrass substitution for integrals. If tanx=t then sin2x=2t/(1+t^2) Is f(t)=2t/(1+t^2) bijective? f'(t)=(2(1+t^2)-2t*2t)/(1+t^2)^2=(2-2t^2)/(1+t^2)^2 so th function has a minimum ar t=-1 and a maximum at t=1 not bijective but we can still try to get an inverse (by restricting the interval) y=2t/(1+t^2) yt^2-2t+y=0 t=(2+-√(4-4y^2))/2y=(1+-√(1-y^2))/y but which variant to pick I can see that lim(y->0) (1+√(1-y^2))/y is indeterminate 2/0 but for (1-√(1-y^2))/y there is 0/0 so i would go with this one
It's multiple choice, so my approach is to eliminate options. x = 0+ and result needs to be 0+. So that eliminates (a). B becomes x / 2 so that's fine. C is also eliminated. So it's B or D. I'd go with B since it looks reasonable. I saved over 5 minutes!
ok when I saw the thumbnail and read "precalculus" I was completely taken aback at the difficulty level of this problem... until I clicked the video and saw it's multiple choice.. then the level dropped drastically.. (physics phd candidate btw)
yes. i did this as well and got +/- in fact, if u analyse bprp's method with more care, u also get +/-. in fact, his method is missing f(0) = 0 sin(2x) = t => x = arcsin(t)/2 needs to be treated with more care since we know this is not always the case. (sine is not one to one) by treating this with more care, we do indeed eventually see the +/- pop out as we saw with the sin(2x) = 2tan(x)/(1+tan^2(x)) method
i’m getting both a and b if i’m using direct trig identities so sin(2x) = 2tanx/1+tan^2x using that, substituting tanx as t and then solving, this gets me both options A and B as correct please explain why?
This is actually not a single valued function. Take, for example, x=π/6. Then sin(2x)=sqrt(3)/2 and tan(x)=1/sqrt(3). Now takex=π/3. Then sin(2x)=sqrt(3)/2 again, but now tan(x)=sqrt(3).
tan(x)=sinx/cosx=(sin(x)*cos(x))/(cos^2(x))=(1/2*sin(2x))/(1/2*(1+cos(2x)))=sin(2x)/(1+sqrt(1-sin^2(2x))) Substitute: x=sin(2x) and tan(x) = f(x) f(x)=x/(1+sqrt(1-x^2)) Another solution, but you can easily show (expanding with the radical conjugate) that this is equivalent to solution B. Had I expanded tan(x) with sin(x) in the first step, I would have directly come up with solution B.
So many trigonometric formulas so so many ways of solving this. My approach was less straightforward (2 candidate functions, only one working) but did not introduce reciprocal functions. It means I focused on double-angle laws (which seemed more instinctive) rather than half-angle ones like you did. Straight away you can use sin2x = 2 tanx/(1+(tanx)²). Now by posing t=tanx you have f(2t/(1+t²))=t. By posing y=2t/(1+t²) you establish that t=(1±√(1-y²))/y so f(y)=(1±√(1-y²))/y; meaning that both A and B are correct? Let's verify: f(sin2x)=(1±cos2x))/sin2x so f(sin2x)=(1+2cos²x-1)/(2sinxcox) [A] or f(sin2x)=(1-1+2sin²x)/(2sinxcosx) [B]. For A this simplifies as f(sin2x)=2cos²x/(2sinxcosx)=cosx/sinx=1/tanx and for B this simplifies as f(sin2x)=2sin²x/(2sinxcosx)=sinx/cosx=tanx so obviously B is the only answer.
ive tried another method which sort of ends up being equivalent to the quadratic method f(sin(2x)) = tan(x) let g be the inverse of f (see I can only use this method if g exists. what if f is not bijective, I would have to use a different method to analyse or try to prove that there cannot exist a non one-to-one f for example...) => sin(2x) = g(tan(x)) t = tan(x) g(t) = 2t/(1+t^2) g is not one-to-one so the inverse splits between options A and B depending on the domain restriction (using quadratic method) A and B together form an inverse relation as well as g(0) = 0, so f(x) = 0 at x = 0 i.e. A, B and f(0) = 0 form an inverse relation. This only works provided f is bijective. (See, we get valid bijective f... but maybe there are functions f out there that are not bijective but i cannot prove why or why not. this is way beyond me).
i guess this is a question of convention, if we decide to make this function relevant to real world application, which f(x) would be more useful in the end. all A, B and even the reciprocals of A and B (in fact the reciprocal of A include x = 0) work. choose a certain domain of f to be bijective... even then, we still get multiple valid f(x) and perhaps the reciprocal of A looks 'prettier' are since it the limit exists in both directions for all 'non end points' and the there is no division by 0. the reciprocal of A is B with a different domain (B does not include 0, reciprocal of A does). yet this is up to interpretation. just A itself is still valid if u decide to make a different domain restriction on g = f^-1
I substituted sin2x as t then used t = sin(2x) = (2tanx)/(1-tan^2 x) you then get a quadratic in tanx which you can solve to get f(t) But I do not know how to tell if the answer is 1 or 2 can anyone help?
Let u=sin2x , so x=.5arcsin(u). Now fuction becomes f(u)=tan(.5arcsin(u)). Apply the formula: .5arcsin(u)= arctan((1-√(1-(u^2)))/u) Now tan cancels arctan, so the function becomes f(u)=(1-√(1-u^2))/u So f(x)=(1-√(1-x^2))/x means B right answer
@@nizamkanchon3311yes, memorising it is really a pain because there are 3 such formulas: .5arcsinx, .5arccosx and .5arctanx. The real pain is the 1st and the 3rd one, cause they r somewhat similar. But if u can remember this formulas, it will save u some time. It is also easy to prove these formulas, whenever i forget i just derive it again😂😂.
tan 2x = 2tan x/(1-tan² x) => tan²x + 2/tan 2x tan x - 1= 0 => tan x = -(1/tan 2x) ± √(1+1/tan²2x) Also tan 2x = sin 2x / cos 2x = sin 2x / √(1 - sin² 2x) So f(sin 2x) = tan x => f(x) = -√(1-x²)/x ± √(1+(1-x²)/x²) = -√(1-x²)/x ± √(1/x²) = (±1 - √(1-x²)) / x So B.
These kind of 'equations' remind me of y' = y as in that we are not looking for a number but another equation such that it will make f(g(x)) = x and we are looking for a g(x) that will make this true..
How to really solve sin(x^2)=sin(x): th-cam.com/video/gCOWo13xxzc/w-d-xo.html
Alternatively, we can use trig identities directly.
tan(x)
= sin(x) / cos(x)
= 2sin²(x) / (2sin(x)cos(x))
= (1 + 2sin²(x) - 1)/(sin(2x))
= (1 - cos(2x))/(sin(2x))
= (1 - √(1 - sin²(2x)) / (sin(2x))
That’s a great idea!
looking at this actually asserts my concern in my original comment
because in fact cos(2x) = +/- sqrt(1-sin^2(2x))
which leads to multiple valid f(x) in the end (both A and B).
In my original comment I even go over where it is possible for f(x) = 0 which both choice A and B miss since the domain of A and B is [-1,0) union (0,1]
@@沈博智-x5y actually both A and B are valid candidates with this approach but using A you end up with f(sin2x)=1/tanx which is wrong :(
@@florianbasier you are missing the fact that cos(2x) can be negative
so
= (1 - cos(2x))/(sin(2x))
= (1 - √(1 - sin²(2x)) / (sin(2x))
are not equivalent
those two lines suggest
cos(2x) = sqrt(1 - sin^2(x)) only
and since sqrt() is non negative, it apparently implies cos(2x) is non negative, which it is not.
(1 + sqrt(1-x^2))/x gives 1/tan(x) ON THE CONDITION that sqrt(1-x^2) = cos(2x), this only works when cos(2x) is non negative..., so when cos(2x) is negative, it's fair game and gives tan(x).
Can someone link the playlist where sir taught these kind of problems anyone ?
i was going to substitute each choice for f(x). there are only 3 numbers anyway
Exactly 😂
What if you don't have enough time for the rest of questions because you have to look deep enough in each option to make sure you are not missing something
@amitanshsrivastav9640 i just have to learn to be fast
This is correct in the range from [-pi/4, pi/4] (except zero, but the limit is correct). It is not correct in general for any real x. The reason this solution fails is that the image of 1/2 arcsin (x) only covers this range, so the substitution does not work in general.
The answer is D because for some values of x the answer would be A and for some others it would be B.
I wish I would have had you as my math teacher way back when I was back in high school way back in the early eighties. I think I learned more from your videos than I ever did sitting in class and struggling through what was being taught in class
sin(2x) = 2sin(x)cos(x) = 2 tan(x)cos²(x) = 2 tan(x)/sec²(x), then write everything in terms of t and invert the expression
To claim that x is the half of arcsin of t, you need to know that x is in [-pi/4, pi/4] which is not in the question. Without that information, there may be other solutions using x=1/2sin^-1(t) + pi for example.
It is especially problematic since the question is framed as an implication. Students have difficulties with inverse trigonometric functions and logic, so it is a disservice to them
This is an interesting problem, but it is not well defined. The answer is not necessarily B. It can also be A depending on what domain values we consider. Remember there is a square root involved. When you take square roots, there is ± depending on the domain/range. Since there is "None of these" in the answer, the correct choice would be D as there is no answer saying "Not enough information given". Thank you for bringing this problem.
Since f(sin(2x))=tanx, I opted to just evaluate each expression at x -> sin(2x) and see which answer simplified to tanx.
No 1/2 angle or inverse identities to worry about that way; not that I don't know them, or couldn't figure them out.
I thought it would be more straightforward in this case.
f(x) = tan[ arcsin(x)/2 ]
there should be some restriction on x when inverting sin(2x)=t.
if we do the multiple choice method of just going through each option one by one then we can see A is valid on some condition (similarly you will find B is valid on some condition)
for A:
f(x) = (1 + sqrt(1-x^2))/x
=> f(sin(2x)) = (1 + sqrt(1 - sin^2(2x)))/sin(2x)
=> f(sin(2x)) = (1 + sqrt(cos^2(2x)))/sin(2x)
=> f(sin(2x)) = (1 + |cos(2x)|)/sin(2x)
=> f(sin(2x)) = (1 - cos(2x))/sin(2x) when cos(2x) f(sin(2x)) = tan(x)
So A is valid provided cos(2x) = 0
Both of these miss f(0) = 0
but if u analyse the limits, B approaches 0 when x approaches 0.
by the very official Multiple Choice Theorem, it's b.) two answers with 1 at the start, two answers with x^1 at the bottom, and two answers subtracting the radical. the only answer containing all 3 is b. unless it's d.), in which case, that stinks.
We can cheat somewhat. consider x so small sin x ≈ x, above zero.
Then a) = 1 + 1- / 0+ = 2- / 0+ = inf
b) = 1 - 1- /0+ = 0-/0- (can't be thrown out, need investigate further)
C) = 0+ - 1- / 0+ = -1±/0+ = inf.
So it's b) or d)
Also I'm on phone, so sorry for notation.
0+ mean as "0 + very small number" and / 0+ = inf means "we so close to zero.dividing gets really big number" - tan(0) is zero, not inf
We can write sin(2x) as 2tanx/ 1+ tan^2x and then take tanx = u so then f(2u/1+u^2) = u then u can find f(x) equals what by using 2u/1+u^2 = x and find u in terms of x so I got 2 values of f(x) but the problem is both are in the options so pls tell why the 1st option is wrong if we use this way
5:05 actually you can use the identity role (cos θ)^2 = 1 - (sin θ)^2, so cos θ =sqrt (1-x) in this case
Nope: cos θ = sqrt( (1-x)(1+x) ) not just sqrt( 1-x )
Cos theta is not equal to sqrt(1-x^2), since cos can be negative, while a square root is not
Well, due to 'just' looking at the situation in a right triangle, the angles for which a cosine is negative are avoided.
But a negative square root? Let's look at complex numbers...
Let's find a complex number z, such that:
Arg(z^(1/2))=pi=Arg(z)/2 => Arg(z)=2pi, but an argument (aka fase) of a complex number needs to be an element of (-pi,pi]
So let's multiply z with 1=e^(-i2pi) and make arg(z) equal to 0, and thus, z is just a positive real number. But Arg(z)=0 also means Arg(z)/2=0 which contradicts Arg(z)/2=pi, so a negative square root indeed doesn't exist. /s
@Apollorion ok, who spoke about triangles here? By considering (right?) triangles, you are restricting yourself to acute angles.
@@TaladrisKpop BlackPenRedPen literally spoke about those right triangles in the video itself. But you're right that I should have named the triangles right triangles and shall correct my message above.
You can also check that, for x-->0, f(0) = f(sin 0) = tan 0 = 0, so 1 and 3 are wrong. Then you substitute sin 2x in B to see if it checks.
Sadly, if you actually check B, it only gives the right value over half the reals. A gives the right value over the other half. Since no answer gives the right value over all reals, the correct answer is D.
If you convert sin2x in terms of tanx and then solve the function, you get a and b
2:25 ive never encountered this type of question. can someone confirm this is correct:
When you say f(sin(2x)) = tan x
And you want to solve f(x)
That means you know the output for the unknown function f(sin(2x)) with x substituted for sin(2x) is the same function as tan(x)
So when you want the function f(x) you need to find a way to rewrite f(sin(2x)) as f(x) by defining sin(2x) as a single variable
So sin(2x) = t
But now you you need to define the other side, tan x to be equal. So you use sin(2x) = t and solve for x, so you can fill that in tan (x) and now you only have t’s since you defined al x’s in terms of t.
And then you replace t with x with what matters is that there is 1 of the same variable on both sides, and the left side is f(x)
Is this correct reasoning?
I tried it the same way bruv.
If we are talking about the domain (-pi/2, pi/2):
Isn't option B true just for angles on the domain [-pi/4, 0) and (0, pi/4],
And then option A is true for angles on the domain (-pi/2, -pi/4] and [pi/4, pi/2)?
I chose D because option B doesn't satisfy the requirement when x is 0 or any other multiple of pi/2.
I'm pretty sure you can prove that x cannot be 0 from the beginning but idk can't come up with it.
You forgot that the input into that f(x) is restricted to sin(2x) in order for the function to equal tan(x). It breaks the function at the correct points to make tan x be correct. When sin(2x) = 0, cos(x) = 0. Your claim is that the resultant function needs to have that domain when it is the composition of the resultant function and sin(2x) does.
@@ronaldking1054 When x=0, sin(2x)=0 and tan(x)=0. Based on the wording of the question, I'm pretty sure that case ought to be handled.
True. The real answer should be x/[1+sqrt(1-x^2)]
@@DoongXiouHuaBeautiful 😀
The answer depends on the x domain we are talking about for x € (-pi/4 +2kpi ,pi/4 +2kpi) U (3pi/4 +2kpi , 5pi/4 +2kpi) where k is any integer k...answer B holds.....but for x € ( pi/4 +2kpi , 3pi/4 +2kpi) U (5pi/4+ 2kpi, 7pi/4 +2kpi) where k is an integer...answer A holds....so for lim x -> pi/2 (-)(left handed limit) ....sin (2x) will approach zero but sill be larger than zero so giving us the correct result of tan x diverging +infinity.... (notice answer B won't work only A will work cause of the domain)
One can actually use the unit circle for this problem and solve geometrically
I love ur videos
tanx=(sin2x)(1+cos2x)
I thought the answr is D.
Because if you put x=0 then f(0)=0 but all the choices diverge for x=0.
I think the domain of the function must be specified
This is almost the Weierstrass substitution for integrals. If tanx=t then sin2x=2t/(1+t^2)
Is f(t)=2t/(1+t^2) bijective?
f'(t)=(2(1+t^2)-2t*2t)/(1+t^2)^2=(2-2t^2)/(1+t^2)^2 so th function has a minimum ar t=-1 and a maximum at t=1 not bijective but we can still try to get an inverse (by restricting the interval)
y=2t/(1+t^2)
yt^2-2t+y=0
t=(2+-√(4-4y^2))/2y=(1+-√(1-y^2))/y but which variant to pick
I can see that lim(y->0) (1+√(1-y^2))/y is indeterminate 2/0 but for
(1-√(1-y^2))/y there is 0/0 so i would go with this one
where can i buy the Identities For You sign?
It's multiple choice, so my approach is to eliminate options. x = 0+ and result needs to be 0+. So that eliminates (a). B becomes x / 2 so that's fine. C is also eliminated. So it's B or D. I'd go with B since it looks reasonable. I saved over 5 minutes!
ok when I saw the thumbnail and read "precalculus" I was completely taken aback at the difficulty level of this problem... until I clicked the video and saw it's multiple choice.. then the level dropped drastically.. (physics phd candidate btw)
Is it legal? This might be th solution on some interval but sin(2x) is NOT injective so we only got some solution....
Berry good!
my brain fr went (2tanx)/(1+tan²x) = sin2x and solved for tanx using quadratic formula. the only issue is that theres a +-
Same here but it's sin2x tbf.
yes.
i did this as well and got +/-
in fact, if u analyse bprp's method with more care, u also get +/-.
in fact, his method is missing f(0) = 0
sin(2x) = t => x = arcsin(t)/2 needs to be treated with more care since we know this is not always the case. (sine is not one to one)
by treating this with more care, we do indeed eventually see the +/- pop out as we saw with the sin(2x) = 2tan(x)/(1+tan^2(x)) method
@@PaadSingh-cr8qg my bad ill fix it
Your videos are always so beneficial, i love trying myself while solving the questions you prepare. Thank you.
i’m getting both a and b if i’m using direct trig identities
so sin(2x) = 2tanx/1+tan^2x
using that, substituting tanx as t and then solving, this gets me both options A and B as correct
please explain why?
Can someone link the playlist where sir taught these kind of problems anyone ?
This is actually not a single valued function. Take, for example, x=π/6. Then sin(2x)=sqrt(3)/2 and tan(x)=1/sqrt(3). Now takex=π/3. Then sin(2x)=sqrt(3)/2 again, but now tan(x)=sqrt(3).
Take x = 0, f(0) = 0, only B gives you zero, so either B or D, and ABC looks very similar to each other, answer should be one of them.
tan(x)=sinx/cosx=(sin(x)*cos(x))/(cos^2(x))=(1/2*sin(2x))/(1/2*(1+cos(2x)))=sin(2x)/(1+sqrt(1-sin^2(2x)))
Substitute: x=sin(2x) and tan(x) = f(x)
f(x)=x/(1+sqrt(1-x^2))
Another solution, but you can easily show (expanding with the radical conjugate) that this is equivalent to solution B.
Had I expanded tan(x) with sin(x) in the first step, I would have directly come up with solution B.
Whome are You adressing, talking about principles I don´t have any memories of? Were we taught these functions at school? (in the 1960-ies?)
I think by a first quick look
f(x) = x/2cos^2(x) ?
no
@ibnSafaa uh u right its not so easy
So many trigonometric formulas so so many ways of solving this. My approach was less straightforward (2 candidate functions, only one working) but did not introduce reciprocal functions. It means I focused on double-angle laws (which seemed more instinctive) rather than half-angle ones like you did. Straight away you can use sin2x = 2 tanx/(1+(tanx)²). Now by posing t=tanx you have f(2t/(1+t²))=t. By posing y=2t/(1+t²) you establish that t=(1±√(1-y²))/y so f(y)=(1±√(1-y²))/y; meaning that both A and B are correct? Let's verify: f(sin2x)=(1±cos2x))/sin2x so f(sin2x)=(1+2cos²x-1)/(2sinxcox) [A] or f(sin2x)=(1-1+2sin²x)/(2sinxcosx) [B]. For A this simplifies as f(sin2x)=2cos²x/(2sinxcosx)=cosx/sinx=1/tanx and for B this simplifies as f(sin2x)=2sin²x/(2sinxcosx)=sinx/cosx=tanx so obviously B is the only answer.
f(x)=tan(0.5arcsin(x)) 🗿
This question is poorly asked, or I'm dumb. How can you have f(sin2x) = f(x)? The x on the left is not equivalent to the one on the right.
對台灣高中生來說 會快速反應 tan2x=2t/(1-t2) t=tanx sin2x= 2t/(1+t2) .....
🥰🥰🥰🥰
ive tried another method which sort of ends up being equivalent to the quadratic method
f(sin(2x)) = tan(x)
let g be the inverse of f (see I can only use this method if g exists. what if f is not bijective, I would have to use a different method to analyse or try to prove that there cannot exist a non one-to-one f for example...)
=> sin(2x) = g(tan(x))
t = tan(x)
g(t) = 2t/(1+t^2)
g is not one-to-one
so the inverse splits between options A and B depending on the domain restriction (using quadratic method)
A and B together form an inverse relation as well as g(0) = 0, so f(x) = 0 at x = 0
i.e. A, B and f(0) = 0 form an inverse relation.
This only works provided f is bijective. (See, we get valid bijective f... but maybe there are functions f out there that are not bijective but i cannot prove why or why not. this is way beyond me).
i guess this is a question of convention, if we decide to make this function relevant to real world application, which f(x) would be more useful in the end.
all A, B and even the reciprocals of A and B (in fact the reciprocal of A include x = 0) work.
choose a certain domain of f to be bijective...
even then, we still get multiple valid f(x) and perhaps the reciprocal of A looks 'prettier' are since it the limit exists in both directions for all 'non end points' and the there is no division by 0.
the reciprocal of A is B with a different domain (B does not include 0, reciprocal of A does).
yet this is up to interpretation.
just A itself is still valid if u decide to make a different domain restriction on g = f^-1
I substituted sin2x as t then used t = sin(2x) = (2tanx)/(1-tan^2 x)
you then get a quadratic in tanx which you can solve to get f(t)
But I do not know how to tell if the answer is 1 or 2
can anyone help?
Let u=sin2x , so x=.5arcsin(u).
Now fuction becomes f(u)=tan(.5arcsin(u)).
Apply the formula:
.5arcsin(u)=
arctan((1-√(1-(u^2)))/u)
Now tan cancels arctan, so the function becomes
f(u)=(1-√(1-u^2))/u
So f(x)=(1-√(1-x^2))/x means B right answer
Good solution. But that trig identity is hard to remember. I often forget it.
@@nizamkanchon3311yes, memorising it is really a pain because there are 3 such formulas: .5arcsinx, .5arccosx and .5arctanx. The real pain is the 1st and the 3rd one, cause they r somewhat similar.
But if u can remember this formulas, it will save u some time. It is also easy to prove these formulas, whenever i forget i just derive it again😂😂.
@avosdelhevia-y7f .5arccosx has 3 formulas.
@@nizamkanchon3311yes, but it is easier than .5arctanx or .5arcsinx. These 2 r almost identical.( only difference is some plus minus).
F(x)=tan(1/2*arcsin(x))
Done, 6 seconds take that
tan 2x = 2tan x/(1-tan² x)
=> tan²x + 2/tan 2x tan x - 1= 0
=> tan x = -(1/tan 2x) ± √(1+1/tan²2x)
Also tan 2x = sin 2x / cos 2x = sin 2x / √(1 - sin² 2x)
So f(sin 2x) = tan x
=> f(x) = -√(1-x²)/x ± √(1+(1-x²)/x²)
= -√(1-x²)/x ± √(1/x²)
= (±1 - √(1-x²)) / x
So B.
I just tried x=0. Option A and C are 2/x and 1/x around zero, where B gives 0/0, making it possible to be correct. So I assumed B.
These kind of 'equations' remind me of y' = y as in that we are not looking for a number but another equation such that it will make f(g(x)) = x and we are looking for a g(x) that will make this true..