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I have two trigonometric equations and I have to find for x and y.

$\sin x \cos y=\frac{1}{4}$ and $3\tan x=\tan y$.

I tried to make it of the form of compound/ allied angles like $\sin(x+y)$ but failed.

Please give me some hint.

Thanks.

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  • $\begingroup$ General solution or in some finite interval? $\endgroup$
    – DonAntonio
    Jan 2 '17 at 11:45
  • $\begingroup$ General solution Don Antonio $\endgroup$ Jan 2 '17 at 11:46
  • $\begingroup$ But give me hint $\endgroup$ Jan 2 '17 at 11:46
  • $\begingroup$ $ 3 \tan x = 3 \frac{\sin x}{\cos x} = \frac{\sin y}{\cos y} = \tan y$. So, $$3 \sin x \cos y = \sin y \cos x$$ Since $\sin x \cos y = \frac{1}{4}$, you obtain $\frac{3}{4}= \sin y \cos x$. Does that help? $\endgroup$
    – YukiJ
    Jan 2 '17 at 11:52
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As you asked for the hint, here is the one:

$\sin x \cos y=\frac{1}{4}$ and $3\tan x=\tan y$.

If you transform the second equation, you willl get:

$$3\tan x=\tan y \implies 3\sin x \cos y-\sin y\cos x=0$$ Using $\sin x \cos y=\frac{1}{4}$, you will get $\sin y\cos x=\frac{3}{4}$ and $\sin x \cos y=\frac{1}{4}$.

$\implies \sin(x+y)=1 $ and $\sin(x-y)=\frac{1}{2}$.

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Hint:

$3\tan x = \tan y$

$\Rightarrow3\sin x\cos y=\cos x\sin y$

$\Rightarrow\cos x\sin y=\frac{3}{4}$

$sin(x+y)=\sin x\cos y+\cos x\sin y =1$

$sin(x-y)=\sin x\cos y-\cos x\sin y =\frac{1}{2}$

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Hint...you can write the second equation in terms of $\sin$ and $\cos$ and use the first equation to obtain $$\cos x\sin y=\frac 34$$

Can you take it from there?

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