@@uniquelegend2711 squaring trigonometric equations produces extra terms which might give u extra solutions, these extra solutions need not satisfy the original equation, so even after squaring the equation , u must individually check if the solutions are satisfying the equation, if not, then reject the solutions.
@@uniquelegend2711when u square the left side it’s (sinx + cosx)^2 then u get the 2 terms squared (which equals 1) plus 2sinxcosx (which is equal sin2x) then u solve 2x = 180 and x = 90
sin(x) + cos(x) = 1 so 1 + sin(2x) = 1 sin(2x) = 0 2x = 0, 2x = 180, 2x = 360, 2x = 540, 2x = 720 x = 0, 90, 180, 270 or 360 180 is invalid and so is 270 so x = 0, x = 180, or x = 360 On 0
Another solution is to multiply both sides to √2/2 √2/2(sinx + cosx) = √2/2 √2/2sinx + √2/2cosx = √2/2 ✍️ as we know, cos45°=sin45°=√2/2 cos45°sinx + sin45°cosx = √2/2 ✍️ and it looks very similar to: sin(α+β)= sinα*cosβ + sinβ*cosα so we rewrite it as sin(45°+x)=√2/2 ✍️ and as we know, if sinx = a, (a ∈ (0;1) ) the solution is: x = (-1)^k*arcsin(a)+πk, k ∈ Z 45°+x = (-1)^k*45+180°*k, k ∈Z x = (-1)^k*45+180°k - 45°, k ∈Z we are given that x ∈[0;360°] k=0, x =45°-45°= 0✅ k = 1, x = -45+180-45=90°✅ k = 2, x = 45+360-45 = 360°✅ k = -1, x = -45-180-45 = -270 ❌ and no need to check anymore, because none will satisfy the given x I guess this is kinda a bit longer, but just wanted to write ^^ hope, I haven't done any mistakes 😅
I really appreciate your comment and method. But this is a specific curriculum math video (International Baccalaureate SL and HL) where the sum formula sin(α+β)= sinα*cosβ + sinβ*cosα is not studied within the IB SL. you can turn sin x +cos x=1 into sin(x+45)=√2/2 or cos(x-45)=√2/2 using a standardized method in Trigonometry: R*sin(x+-a)or R*cos(x+-a), especially if the question was harder.
You can easily deduce that this can only happen at the right angles (0, pi/2, pi, 3pi/2) through case analysis of the quadrants. In the first quadrant (0, pi/2) the sum is always greater than 1 because the sum of two side lengths of a triangle is always greater than the length of the other side. In the second and fourth quadrants (pi/2, pi) and (3pi/2, 2pi) the sum is always less than 1 because one (either cos(x) or sin(x)) is negative and the other is positive but less than 1. Finally, in the third quadrant the sum is negative. Knowing this, we just need to try these 4 angles. After testing, we can conclude that x = 0, pi/2 and all equivalent angles modulo 2pi.
I approached the question like this, 1-cosx = 2sin²x/2 & sinx = 2sinx/2cosx/2, divide both sides by sinx/2 and the equation becomes 2sinx/2 = 2cosx/2, so tan x/2 = nπ+pie/4 & sinx = 0 which is nπ.
sin ( x) + cos ( x) = 1 √ 2 * cos ( x - π /4) =.1 cos ( x - π /4) = 1/√2 = cos (π /4) x - π /4 = 2 n π + π /4, 2 n π - π /4 x = 2 n π + π/2, 2 n π, for imtegral n x = π /2, 0, 2π, 0 are only solution in (0, π)
0:19 360 is not the same as 360 DEGREES. If you only write 360, you don't say anything that it is degrees you mean. You think it is 360 radians, 64800 degrees.
The 2x = 0 you got is one possibility, but it's not the only possibility. To get all the possibilities, you've should have written: _From sin(2x) = 0 get 2x = 180 n for some integer n._ Continuing on you'd write: Thus x = 90 n for some integer n. And because 0
2 sin(x) cos(x) is just sin(2x) !!! Much easier to solve this way!
Don’t need to watch. Just square both sides!!!
No, you create extraneous solutions by squaring. Use the harmonic form instead.
My sir told square would create false solution
Plz can you tell what is that
@@uniquelegend2711 squaring trigonometric equations produces extra terms which might give u extra solutions, these extra solutions need not satisfy the original equation, so even after squaring the equation , u must individually check if the solutions are satisfying the equation, if not, then reject the solutions.
@@uniquelegend2711when u square the left side it’s (sinx + cosx)^2 then u get the 2 terms squared (which equals 1) plus 2sinxcosx (which is equal sin2x) then u solve 2x = 180 and x = 90
@@uniquelegend2711 it will add pi at least which is not a solution
cos(x)=(e^ix+e^-ix)/2
sin(x)=-i(e^ix-e^-ix)/2
(1-i)e^ix+(1+i)e^-ix=2
1-i=sqrt(2)e^-ipi/4
1+i=sqrt(2)e^ipi/4
e^i(x-pi/4)+e^-i(x-pi/4)=sqrt(2)
cos(x-45°)=1/sqrt(2)
x-45°={-45°, 45°, 315°}
x={0°, 90°, 360°}
sin(x) + cos(x) = 1
so 1 + sin(2x) = 1
sin(2x) = 0
2x = 0, 2x = 180, 2x = 360, 2x = 540, 2x = 720
x = 0, 90, 180, 270 or 360
180 is invalid and so is 270
so
x = 0, x = 180, or x = 360
On 0
Another solution is to multiply both sides to √2/2
√2/2(sinx + cosx) = √2/2
√2/2sinx + √2/2cosx = √2/2
✍️ as we know, cos45°=sin45°=√2/2
cos45°sinx + sin45°cosx = √2/2
✍️ and it looks very similar to:
sin(α+β)= sinα*cosβ + sinβ*cosα
so we rewrite it as sin(45°+x)=√2/2
✍️ and as we know, if sinx = a, (a ∈ (0;1) ) the solution is:
x = (-1)^k*arcsin(a)+πk, k ∈ Z
45°+x = (-1)^k*45+180°*k, k ∈Z
x = (-1)^k*45+180°k - 45°, k ∈Z
we are given that x ∈[0;360°]
k=0, x =45°-45°= 0✅
k = 1, x = -45+180-45=90°✅
k = 2, x = 45+360-45 = 360°✅
k = -1, x = -45-180-45 = -270 ❌
and no need to check anymore, because none will satisfy the given x
I guess this is kinda a bit longer, but just wanted to write ^^ hope, I haven't done any mistakes 😅
I really appreciate your comment and method.
But this is a specific curriculum math video (International Baccalaureate SL and HL) where the sum formula sin(α+β)= sinα*cosβ + sinβ*cosα is not studied within the IB SL.
you can turn sin x +cos x=1 into sin(x+45)=√2/2 or cos(x-45)=√2/2 using a standardized method in Trigonometry: R*sin(x+-a)or R*cos(x+-a), especially if the question was harder.
@@ibmathmasterWhat's sl and hl?
IB math is an international school curriculum for grades 11 and 12, standard level SL or high level HL
You can easily deduce that this can only happen at the right angles (0, pi/2, pi, 3pi/2) through case analysis of the quadrants. In the first quadrant (0, pi/2) the sum is always greater than 1 because the sum of two side lengths of a triangle is always greater than the length of the other side. In the second and fourth quadrants (pi/2, pi) and (3pi/2, 2pi) the sum is always less than 1 because one (either cos(x) or sin(x)) is negative and the other is positive but less than 1. Finally, in the third quadrant the sum is negative.
Knowing this, we just need to try these 4 angles. After testing, we can conclude that x = 0, pi/2 and all equivalent angles modulo 2pi.
I approached the question like this, 1-cosx = 2sin²x/2 & sinx = 2sinx/2cosx/2, divide both sides by sinx/2 and the equation becomes 2sinx/2 = 2cosx/2, so tan x/2 = nπ+pie/4 & sinx = 0 which is nπ.
Good explanation at end of the video.
Glad you liked it
sin(x) + cos(x) = 1
(sin(x) + cos(x))^2 = 1^2 = 1
sin^2(x) + 2*sin(x)*cos(x) + cos^2(x) = 1
1 + 2*sin(x)*cos(x) = 1
2*sin(x)*cos(x) = 0
sin(2*x) = 0
2*x = k*pi, integer k
x = (k/2)*pi, integer k
sinx + cosx = 1
(sinx + cosx)^2 = 1^2 = 1
2(sinx)(cosx) = 0
sinx = 0, cosx = 1
or
cosx = 0, sinx = 1
x = 2nπ, (4n + 1)π/2
sin x + cos x = 1
Sqrt 2*sin (x+Pi/4)=1
Sin(x+Pi/4)=1/sqrt 2
x+Pi/4=Pi/4+2nPi
x=0 or 2Pi
x= Pi/2 (0
sin ( x) + cos ( x) = 1
√ 2 * cos ( x - π /4) =.1
cos ( x - π /4) = 1/√2 = cos (π /4)
x - π /4 = 2 n π + π /4, 2 n π - π /4
x = 2 n π + π/2, 2 n π, for imtegral n
x = π /2, 0, 2π, 0 are only solution in (0, π)
sinx+tg45*cosx = 1 , multiply both sides by cos45 .... you get sin(x + 45) = cos45 .....
2nπ+π/4 & 2nπ.
90
90, 360
+/- kpi/2
X= 0, X= π/2. Tiene infinitas soluciones.
Para qué elevar al ⬛.
0:19 360 is not the same as 360 DEGREES. If you only write 360, you don't say anything that it is degrees you mean. You think it is 360 radians, 64800 degrees.
It’s just a domain bro it’s like a given that when you do that you use degrees cause idk most things are in the domain of [0, 2pi]
By Squaring both sides
1+sin2x=1
Sin2x=0
Sin2x=sin0
2x=0
X=0 degree
The 2x = 0 you got is one possibility, but it's not the only possibility.
To get all the possibilities, you've should have written:
_From sin(2x) = 0 get 2x = 180 n for some integer n._
Continuing on you'd write:
Thus x = 90 n for some integer n.
And because 0