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I'm trying to solve this problem:

If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \dfrac{\theta-\phi}2$.

So seeing $\dfrac{\theta-\phi}2$ in the argument of the tangent function, I first thought of converting the left-hand sides of the givens to products which gave me: $$2\sin\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=a\quad,\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b$$

But then, on dividing the two equations (assuming $b\ne0$), I just get the value of $\tan\dfrac{\theta+\phi}2$.

I don't know how else to proceed. Any help would be appreciated!

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up vote 6 down vote accepted

Method $1:$

Squaring & adding what you have derived $$4\cos^2\frac{\theta-\phi}2=a^2+b^2$$

$$\implies \sec^2\frac{\theta-\phi}2=\frac4{a^2+b^2}$$

$$\implies \tan^2\frac{\theta-\phi}2=\frac4{a^2+b^2}-1=\frac{4-a^2-b^2}{a^2+b^2}$$

Method $2:$

$$\text{As }\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b,$$

$$\implies \sec\frac{\theta-\phi}2=\frac2{b\sec\frac{\theta+\phi}2}$$

$$\implies \sec^2\frac{\theta-\phi}2=\frac4{b^2\sec^2\frac{\theta+\phi}2} =\frac4{b^2\left(1+\tan^2\frac{\theta-\phi}2\right)}=\frac4{b^2\left(1+\frac{a^2}{b^2}\right)}$$ as $\tan\frac{\theta+\phi}2=\frac ab$

$$\implies \sec^2\frac{\theta-\phi}2=\frac4{b^2+a^2}$$

$$\text{Now, }\tan^2\frac{\theta-\phi}2=\sec^2\frac{\theta-\phi}2-1$$

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It may be a long way to find it by mine, but it works. As $$\sin\theta+\sin\phi=a,~~~\cos\theta+ \cos\phi=b$$ then $$\cos(\theta-\phi)=\cos\phi\cos\theta+\sin\phi\sin\theta=\frac{a^2+b^2-2}2$$ Now use: $$\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$

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(+1) Nice solution. – Mhenni Benghorbal Jul 27 '13 at 12:11

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