Question

If $\frac{\sin 4\theta +\sin 5\theta }{2\cos 3\theta +1}=\sin \left(m\theta \right)+\sin \left(n\theta \right)$ where $m,n\in N$, then $(m+n)$ equals ?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given,

$\frac{sin4\theta +sin5\theta }{2cos\theta +1}=sin\left(m\theta \right)+sin\left(n\theta \right)$

Taking LHS
$\frac{2sin\left(\frac{4\theta +5\theta }{2}\right)cos\left(\frac{5\theta -4\theta }{2}\right)}{2(1-2si{n}^2(\frac{3\theta }{2})+1}$

$⟹\frac{2sin\left(\frac{9\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}{3-4si{n}^2\left(\frac{3\theta }{2}\right)}$

Multiply Numerator and denominater by $sin\left(\frac{3\theta }{2}\right)$

$⟹\frac{2sin\left(\frac{9\theta }{2}\right)cos\left(\frac{\theta }{2}\right)sin\left(\frac{3\theta }{2}\right)}{3sin\left(\frac{3\theta }{2}\right)-4si{n}^3\left(\frac{3\theta }{2}\right)}$

$⟹\frac{2sin\left(\frac{9\theta }{2}\right)cos\left(\frac{\theta }{2}\right)sin\left(\frac{3\theta }{2}\right)}{sin\left(\frac{9\theta }{2}\right)}$

$⟹2sin\left(\frac{3\theta }{2}\right)cos\left(\frac{\theta }{2}\right)$

$⟹sin\left(\frac{3\theta +\theta }{2}\right)+sin\left(\frac{3\theta -\theta }{2}\right)$

$⟹sin2\theta +sin\theta$

On comparing it with RHS

we get $m=2,n=1$

So the value of $m+n=2+1=3$