Trigonometry #156

Geometry Level 4

Find the smallest positive number p p for which the equation cos ⁡ ( p sin ⁡ x ) = sin ⁡ ( p cos ⁡ x ) \cos(p\sin x)=\sin(p\cos x) has a solution for x ∈ [ 0 , 2 π ] x\in[0,2\pi] .

Enter the value of π 2 p 2 \frac{\pi^{2}}{p^{2}} .

This problem is part of the set Trigonometry .


The answer is 8.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Thushar Mn
Feb 28, 2015

psinx+ pcosx =pi/2 so , p=pi/2(sinx+cosx)

max value of sinx +cosx =root 2. so min value of p=pi/(2 root 2).

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C o s ( p S i n X ) = S i n ( p C o s X ) = C o s ( π 2 − p C o s X ) . ⟹ p S i n X = π 2 − p C o s X p ( S i n X + C o s X ) = π 2 . . . . ( 1 ) L e t f ( X ) = S i n X + C o s X , F o r e x t r e m e v a l u e s , f ′ ( X ) = C o s X − S i n X = 0 , ⟹ , S i n X = C o s X , ∴ X = π 4 . ∴ f r o m ( 1 ) p ( S i n X + C o s X ) = p ∗ ( 1 2 + 1 2 ) = p ∗ 2 = π 2 . ⟹ p = π 2 ∗ 2 ∴ π 2 p 2 = 8 Cos(pSinX)=Sin(pCosX)=Cos(\dfrac \pi 2 -pCosX).\\\implies~pSinX= \dfrac \pi 2 -pCosX \\p(SinX+CosX)=\dfrac \pi 2 ....(1)\\ Let~~f(X) = SinX + CosX, ~~~~~ For~extreme ~values,\\f'(X)=CosX - SinX=0,~~\implies, SinX=CosX,~~\therefore~X= \dfrac \pi 4 .\\\therefore~from~(1)~~p(SinX+CosX)=p*(\dfrac 1 {\sqrt 2} + \dfrac 1 {\sqrt 2} )=p*\sqrt 2=\dfrac \pi 2.\\ \implies~p=\dfrac \pi {2*\sqrt 2 }~~\therefore ~\dfrac {\pi^2}{p^2}= ~~\Large \color{#D61F06}{8}

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