Question

$\cos \alpha +\cos \beta =a,\sin \alpha +\sin \beta =b,\alpha -\beta =2\theta$. Find the value of $\frac{\cos 3\theta }{\cos \theta }$.

• A. $3-a^2-b^2$
• B. $\frac{a^2+b^2}{4}$
• C. $a^2+b^2-1$
• D. $a^2+b^2-3$

Answer 1

Answer: (D)

$a^2+b^2-3$

$\alpha -\beta =2\theta$
$\cos \alpha +\cos \beta =a$ ...(1)
$\sin \alpha +\sin \beta =b$ ...(2)
Squaring and adding eq. (1) & eq. (2)
${\cos }^2\alpha +{\cos }^2\beta +{\sin }^2\alpha +{\sin }^2\beta +2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=a^2+b^2\Rightarrow 1+1+2\cos (\alpha -\beta )=a^2+b^2\Rightarrow 2\cos (\alpha -\beta )=a^2+b^2-2\Rightarrow 2\cos 2\theta =a^2+b^2-2$
$\frac{\cos 3\theta }{\cos \theta }=\frac{4{\cos }^3\theta -3\cos \theta }{\cos \theta }=4{\cos }^2\theta -3\Rightarrow \frac{\cos 3\theta }{\cos \theta }=2(1+\cos 2\theta )-3=2\cos 2\theta -1\Rightarrow \frac{\cos 3\theta }{\cos \theta }=(a^2+b^2-2)-1=a^2+b^2-3$
Hence, option 'D' is correct.