Trigonometry Help

Hey, can anybody provide a solution for this problem? I am totally clueless.

Q.)Find the possible real values of (x,y)(x,y) satisfying sinx+cosx=y22y+4y2+2y+4.\sin { x } +\cos { x } =\frac { { y }^{ 2 }-2y+4 }{ { y }^{ 2 }+2y+4 }.

Note by Rishi Sharma
3 years, 1 month ago

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sinx+cosx[2,2]\sin x +\cos x \in [-\sqrt{2},\sqrt{2}]

The range of the right side is [13,3] \left[ \frac{1}{3},3 \right]

So, sinx+cosx13\sin x +\cos x ≥\frac{1}{3}

Now replacing sinx=2t1+t2,cosx=1t21+t2\sin x =\frac{2t}{1+t^2},\quad \cos x =\frac{1-t^2}{1+t^2}

Where t=tanxt=\tan x, we will then get,

2t23t10t[3174,3+174]x[nπ+tan1(3174),nπ+tan1(3+174)]2t^2-3t-1≤0 \\ t \in \left[\frac{3-\sqrt{17} }{4}, \frac{3+\sqrt{17}}{4} \right] \\ \therefore x \in \left[ n\pi+\tan^{-1}\left( \frac{3-\sqrt{17}}{4}\right) , n\pi+\tan^{-1}\left( \frac{3+\sqrt{17}}{4}\right)\right]

Akshat Sharda - 3 years, 1 month ago

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Exactly what @Deeparaj Bhat said... sinx+cosx=y22y+4y2+2y+4=k\sin { x } +\cos { x } =\frac { { y }^{ 2 }-2y+4 }{ { y }^{ 2 }+2y+4 }=k

k=y22y+4y2+2y+4k=\dfrac{y^2-2y+4}{y^2+2y+4}     y2(k1)2y(k+1)+4(k1)=0....(1)\implies y^2(k-1)-2y(k+1)+4(k-1)=0....(1) Since yRy\in R the discriminate of quadratic must be non-negative.     (k+1)24(k1)20\implies (k+1)^2-4(k-1)^2\ge 0     (3k1)(k3)0    k[13,3]\implies (3k-1)(k-3)\le 0\implies k\in[\dfrac 13,3]

While sinx+cosx[2,2]\sin x+\cos x\in[-\sqrt2,\sqrt 2]

Hence k[13,2]\boxed{k\in[\dfrac 13,\sqrt 2]}

Solve 11 using quadratic formula:

y=1+k±(k3)(3k1)1k,k1\boxed{y=\dfrac { 1+k\pm \sqrt { \left( k-3 \right) \left( 3k-1 \right) } }{ 1-k } },k\ne 1

While solving sinx+cosx=k\sin x+\cos x=k

sin(x+π4)=k2\sin \left(x+\dfrac{\pi}{4}\right)=\dfrac k{\sqrt 2}

    x=π4+2nπ±sin1(k2)\implies x=-\dfrac{\pi}{4}+2n\pi\pm\sin^{-1}\left(\dfrac{k}{\sqrt 2}\right)

While when k=1k=1, we get y=0y=0 while x=0,π/2,2π,5π2x=0,\pi/2,2\pi,\dfrac{5\pi}2\cdots

Rishabh Jain - 3 years, 1 month ago

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Let sinx+cosx=k

It is simple to show that k>= -sqrt(2)

Now, k(y^2+2y+4)=y^2 -2y +4

y^2 (k-1) +2y(k+1) +4(k-1) = 0

For y to be real, D >= 0

This implies 1/3 >= k >= -sqrt(2)

Thus for all real values of x such that k<=1/3 , we will get two real values of y.

I can't solve it further, any help appreciated.

Sorry for no latex.

Harsh Shrivastava - 3 years, 1 month ago

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@Deeparaj Bhat @Rishabh Cool @Prakhar Bindal please help

Rishi Sharma - 3 years, 1 month ago

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@Akshat Sharda the answer is not correct.

Rishi Sharma - 3 years, 1 month ago

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Can you please elaborate what I've done wrong?

Akshat Sharda - 3 years, 1 month ago

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Even I am confused. First you also need to find the value of y as well. And lastly the correct answer (from a credible source) is x=2nπ+cosπ4±cos1k2y=1+k±(k3)(3k1)1k;k[13,2]{1}x=2n\pi +\cos { \frac { \pi }{ 4 } \pm } \cos ^{ -1 }{ \frac { k }{ \sqrt { 2 } } } \\ y=\frac { 1+k\pm \sqrt { \left( k-3 \right) \left( 3k-1 \right) } }{ 1-k } \quad ;\quad k\in \left[ \frac { 1 }{ 3 } ,\sqrt { 2 } \right] -\left\{ 1 \right\}

Rishi Sharma - 3 years, 1 month ago

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@Rishi Sharma In the expression for xx, it should've been π4\frac{\pi}{4} instead of its cosine and for k=1k=1, we have y=0y=0.

Now, coming to how to solve it. Put the RHS as kk. Note that k2k \leq \sqrt{2} due to the LHS.

Also, via calculus (or inequalities) you can see that 13k\frac{1}{3} \leq k (as yy is real).

Now, finding xx in terms of kk is easy.

Also, for k1k \neq 1 we can solve for yy as a quadratic. If k=1k=1, then y=0y=0.

And we're done.

Deeparaj Bhat - 3 years, 1 month ago

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@Deeparaj Bhat Thanks. Got it

Rishi Sharma - 3 years, 1 month ago

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