Question

Proof:
cos 3 theta sin 2 theta - cos 4 theta sin theta= (cos 2 theta)/ cosec theta​

Answer 1

Topic :-

Trigonometry

To Prove :-

$\cos3\theta\sin2\theta-\cos4\theta\sin\theta=\dfrac{\cos2\theta}{\csc\theta}$

Proof :-

Solving LHS,

$\cos3\theta\sin2\theta-\cos4\theta\sin\theta$

Multiply and Divide by 2,

$\dfrac{2}{2}\left[\cos3\theta\sin2\theta-\cos4\theta\sin\theta\right]$

$\dfrac{1}{2}\left[2\cos3\theta\sin2\theta-2\cos4\theta\sin\theta\right]$

Using 2cosAsinB = sin(A + B) - sin(A - B),

$\dfrac{1}{2}\left[\{\sin(3\theta+2\theta)-\sin(3\theta-2\theta)\}-\{\sin(4\theta+\theta)-\sin(4\theta-\theta)\}\right]$

$\dfrac{1}{2}\left[\{\sin(5\theta)-\sin(\theta)\}-\{\sin(5\theta)-\sin(3\theta)\}\right]$

$\dfrac{1}{2}\left[\sin(5\theta)-\sin(\theta)-\sin(5\theta)+\sin(3\theta)\right]$

$\dfrac{1}{2}\left[\sin(5\theta)-\sin(5\theta)+\sin(3\theta)-\sin(\theta)\right]$

$\dfrac{1}{2}\left[0+\sin(3\theta)-\sin(\theta)\right]$

$\dfrac{1}{2}\left[\sin(3\theta)-\sin(\theta)\right]$

$\bold{Using\:\sin A-\sin B=2\cos\left( \dfrac{A+B}{2}\right)\sin\left( \dfrac{A-B}{2}\right),}$

$\dfrac{1}{2}\left[2\cos\left( \dfrac{3\theta+\theta}{2}\right)\sin\left(\dfrac{3\theta-\theta}{2}\right)\right]$

$\dfrac{1}{\not{2}}\left[\not{2}\cos\left( \dfrac{4\theta}{2}\right)\sin\left(\dfrac{2\theta}{2}\right)\right]$

$\left[\cos\left( 2\theta\right)\sin\left(\theta\right)\right]$

$\cos\left( 2\theta\right)\cdot\dfrac{1}{\csc\left(\theta\right)}$

$\left(\because \sin\theta=\dfrac{1}{\csc\theta}\right)$

$\dfrac{\cos2\theta}{\csc\theta}$

RHS,

$\dfrac{\cos 2\theta}{\csc\theta}$

We observe that LHS = RHS.

Hence, Proved !!