Question

If $sin\alpha +sin\beta =aandcos\alpha +cos\beta =b$, prove that

(i) $sin(\alpha +\beta )=\frac{2ab}{a^2+b^2}$

(ii) $cos(\alpha -\beta )=\frac{a^2+b^2-2}{2}$

Answer 1

(i) We have,

$sin\alpha +sin\beta =aandcos\alpha +cos\beta =b$ . . .(A)

Squaring and adding, we get

$si{n}^2\alpha +si{n}^2\beta +2sin\alpha sin\beta +co{s}^2\alpha +co{s}^2\beta +2cosacos\beta =a^2+b^2\Rightarrow 1+1+2(sin\alpha sin\beta +cos\alpha cos\beta )=a^2+b^2\Rightarrow 2(sin\alpha sin\beta +cos\alpha cos\beta )=a^2+b^2$

-2

$\therefore 2cos(\alpha -\beta )=a^2+b^2-2$

Thus, $cos(\alpha -\beta )=\frac{a^2+b^2-2}{2}$ . . .(ii)

Again,

$sin\alpha +sin\beta =a\Rightarrow 2sin\frac{\alpha +\beta }{2}.cos\frac{\alpha -\beta }{2}=acos\alpha +cos\beta =b\Rightarrow 2cos\frac{\alpha +\beta }{2}.cos\frac{\alpha -\beta }{2}=b$
$\Rightarrow tan\frac{\alpha +\beta }{2}=\frac{a}{b}$ . . .(B)

Now,

$sin(\alpha +\beta )=\frac{2tan\frac{\alpha +\beta }{2}}{1+ta{n}^2\left(\frac{\alpha +\beta }{2}\right)}=\frac{2\frac{a}{b}}{1+\frac{a^2}{b^2}}=\frac{2ab}{a^2+b^2}$

Thus,

$sin(\alpha +\beta )=\frac{2ab}{a^2+b^2}$

(ii) $cos(\alpha -\beta )=\frac{a^2+b^2-2}{2}$

We have,

$sin\alpha +sin\beta =aandcos\alpha +cos\beta =b$

Squaring and adding, we get

$si{n}^2\alpha +si{n}^2\beta +2sin\alpha sin\beta +co{s}^2\alpha +co{s}^2\beta +2cos\alpha cos\beta =a^2+b^2\Rightarrow 1+1+2(sin\alpha sin\beta +cos\alpha cos\beta )=a^2+b^2\Rightarrow 2(sin\alpha sin\beta +cos\alpha cos\beta )=a^2+b^2\therefore 2cos(\alpha -\beta )=a^2+b^2-2\text{Thus},cos(\alpha -\beta )=\frac{a^2+b^2-2}{2}$