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I am studying for my trigonometry examination. But I cannot figure out how to this sum from the chapter "Compound and Multiple Angles". I am in the eleventh standard. The sum goes like this :

$\sin x + \sin y = a$ and $\cos x+\cos y=b$. Find $\tan \frac{x+y}{2}$

Any help will be appreciated.

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1 Answer 1

up vote 2 down vote accepted

Recall the two (well, basically one) identities for the sine of a sum and the sine of a difference. We have $$\sin(a+b)=\sin a\cos b+\cos a\sin b,$$ and $$\sin(a-b)=\sin a\cos b-\cos a\sin b.$$ Adding we get $$\sin(a+b)+\sin(a-b)=2\sin a \cos b.$$ Put $x=a+b$ and $y=a-b$. Then $a=\frac{x+y}{2}$ and $b=\frac{x-y}{2}$. It follows that $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).\tag{1}$$

Use the addition/subtraction laws for cosine to get a similar expression for $\cos x+\cos y$, and it will be almost over.

Remark: The identity (1) may already be part of your list of standard identities, and so may the corresponding identity about $\cos x+\cos y$. In that case, the calculation of $\tan\left(\frac{x+y}{2}\right)$ in terms of $a$ and $b$ can be done in one line.

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