Question

If $\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$, then

• A. $\cos \frac{\theta -\varphi }{2}=\pm \frac{1}{2}\sqrt{a^2+b^2}$
• B. $\cos \frac{\theta -\varphi }{2}=\pm \frac{1}{2}\sqrt{a^2-b^2}$
• C. $\tan \frac{\theta -\varphi }{2}=\pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$
• D. $\cos (\theta -\varphi )=\frac{a^2+b^2-2}{2}$

Answer 1

Answer: (A), (C), (D)

$\cos \frac{\theta -\varphi }{2}=\pm \frac{1}{2}\sqrt{a^2+b^2}$
$\cos (\theta -\varphi )=\frac{a^2+b^2-2}{2}$
$\tan \frac{\theta -\varphi }{2}=\pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$

$\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$
Squaring both equations we get
${\sin }^2\theta +{\sin }^2\varphi +2\sin \theta \sin \varphi =a^2$ ....... $\left(i\right)$
${\cos }^2\theta +{\cos }^2\varphi +2\cos \theta \cos \varphi =b^2$ ....... $\left(ii\right)$
Adding equations $\left(i\right)$ and $\left(ii\right)$, we get
$({\sin }^2\theta +{\cos }^2\theta )+({\sin }^2\varphi +{\cos }^2\varphi )+2\sin \theta \sin \varphi +2\cos \theta \cos \varphi =a^2+b^2$
$2+2[\cos \theta \cos \varphi +\sin \theta \sin \varphi ]=a^2+b^2$

$2+2\left[\cos (\theta -\varphi )\right]=a^2+b^2$ ....... $\left(iii\right)$
$2+2\left[2{\cos }^2\left(\frac{\theta -\varphi }{2}\right)-1\right]=a^2+b^2[\because \cos 2\theta =2{\cos }^2\theta -1]$
$4{\cos }^2\left(\frac{\theta -\varphi }{2}\right)=a^2+b^2$
$\cos \left(\frac{\theta -\varphi }{2}\right)=\pm \frac{1}{2}\sqrt{a^2+b^2}$ which is option (A)
Using equation $\left(iii\right)$
$2+2\cos (\theta -\varphi )=a^2+b^2$
$\cos (\theta -\varphi )=\frac{a^2+b^2-2}{2}$ which is option (D)
${\sin }^2\left(\frac{\theta -\varphi }{2}\right)=1-{\cos }^2\left(\frac{\theta -\varphi }{2}\right)=1-\frac{1}{4}(a^2+b^2)$
$\sin \left(\frac{\theta -\varphi }{2}\right)=\pm \frac{1}{2}\sqrt{4-a^2-b^2}$
$\tan \left(\frac{\theta -\varphi }{2}\right)=\frac{\sin \left(\frac{\theta -\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}=\frac{\pm \frac{1}{2}\sqrt{4-a^2-b^2}}{\pm \frac{1}{2}\sqrt{a^2+b^2}}$
$\tan \left(\frac{\theta -\varphi }{2}\right)=\pm \sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$ which is option (C)