Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 4 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 $\tan\frac{\theta+\phi}2=\frac ab$

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

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

share|improve this answer

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: $$\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$

share|improve this answer
    
(+1) Nice solution. –  Mhenni Benghorbal Jul 27 '13 at 12:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.