Star Problem # 9, 10 from Trigonometry | Problem Series for JEE Aspirants
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Sir can you please make a one shot video for advanced trigonometry questions for class 11.
Please sir
1st problem by integration can be solved 10:19
1st problem can be done using simpler way, open bracket, then the equation is in the form acosx+bsinx, and it's min value is √a^2+b^2
Can u elaborate plzz.
bhai likh kar bhej mujhe nhi dikha rha yeh bante huye
@@shivx3295 exactly
f(x) = (Cox)sinx+(√sin^2x+ sin^2alpha)cosx
a=cosx
b=√(sin^x+sin^2alpha)
@@fictionaldino3733 🤣🤣
Sir pyq wala batch 11th ke liye bhi hai kya
By differentiation sir , for f'(x)=0
f(x )= cotx
And cot x =+- ✓(1+sin² a) with some manipulations of trigonometry.
I have an interesting solution for first question
consider two vectors
v₁=(sinx)î+(cosx)ĵ
v₂=(cosx)î+(sqrt(sin^2(x)+sin^2(α)))ĵ
now, dot product:
v₁·v₂=|v₁||v₂|cosθ
cosθ=(v₁·v₂)/(|v₁||v₂|)
range of cosθ is [-1,1]. so,
-1
Isse acha toh seedha cauchy lagade
sir make a video on Ramanujan nested radical problem
Sir please reappload the lecture of application of derivatives
Sir TanA + TanB +TanC wale q mei Jensen's Inequality se easily ho sakta hai
sir, lega sir ki nda playlist aadhuri q hai ??? algebra ke lectures nhi mil rha
can be solved much easily by cs inequality .
👍👍
Apke function ch ke lec hide ho gaye kese dhekhe plz
👍👍👍👍👍
Sir where is the playlist of ITF
Sir cauchy schuartz
In 1st problem y =cos(-----) can't we say that range of y = range of cos =[-1,1]
This is cosx ( ) not in angle
It is cosx multiplied not cos (__)
Sorry for misunderstanding
@@parthtomar6987 no worries I also saw what u did but later I realised
2nd question gensen inequality lagake oral kardiya
Sir badal gaye hai.
Wo hamare comments nhi padhte hai
He ignores us 😢
Alpha type nhi kr skta sir,,to (alpha= y )kr diya hu😊..
Baki bahut pyara angreji aur simple language ka prayog hai,,plz have a look!!😊👍
Solution:-👇
The range of F(x) can be found
by analyzing the expression
sin^2(x) + sin^2(y) ≥ 0 (since sine squared is always non-negative ; sin^2 x ≥0)
So, the expression inside the square root is always non-negative.👍
The maximum value of sin^2(x) + sin^2(y) occurs when x = y = π/2:
sin^2(π/2) + sin^2(π/2) = 1 + 1 = 2
So, the maximum value of the expression inside the square root is 2.
Now, let's analyze the function F(x):
F(x) = cos x [sin x + √(sin^2(x) + sin^2(y))]
Since cos x and sin x are bounded between -1 and 1;[-1,1]
and the square root expression is non-negative, the maximum value of F(x) occurs when cos x = 1 and sin x = 1:
F(max) = 1 [1 + √2]
The minimum value of F(x) occurs when cos x = -1 and sin x = -1:
F(min) = -1 [-1 + √2]
So, the range of F(x) is:
[-1 + √2, 1 + √2]
Thankyou to read!
Radhe radhe🙏
you can't put sin x =1 and cos x =1 for the same value of X
At sin x = 1 cos x = 0 and vise versa. :)
@@HobinderSinghA yes, your cross statement is valid bro! But here we don't need to comment on exact value at any instance,, instead we just have to analyze the boundaries, so we can overlook that fact!🙂
And practically,, bro As JEE aspirants perspective you don't need to write each steps na,, just understand the core approach!
fundamentally so wrong you cannot just do even if this work in some question but it will not work in all so do not do that
@@shivx3295 hmm make sense!👍
First question i solved through differentiation
Can you show me bro
Sir you're energy is going down
What happened
Is something wrong 🙁
sir is taking multiple batches
Sir please give me answer of this question.
Are you muslim?