Question

Prove the following trigonometric equation:
$\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x=0$

Answer 1

Hint:We can see in the L.H.S of the given equation, sine and cosine are in the form of $\sin C+\sin D$ and $\cos C-\cos D$ respectively. The identities of sine and cosine that we have just mentioned then apply in $\sin 3x+\sin x$ and $\cos 3x-\cos x$ followed by simplification.

Complete step-by-step answer:
The equation given in the question that we have to prove is:
$\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x=0$
We are going to simplify the L.H.S of the above equation.
In the above equation, the trigonometric expression $\sin 3x+\sin x$ is in the form of $\sin C+\sin D$. So, applying the identity of $\sin C+\sin D$ in $\sin 3x+\sin x$ we get,
$\sin C+\sin D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\sin 3x+\sin x=2\sin 2x\cos x$
It is also visible that the trigonometric expression $\cos 3x-\cos x$ is in the form of $\cos C-\cos D$. So, applying the identity of $\cos C-\cos D$ in $\cos 3x-\cos x$ we get,
$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$
$\cos 3x-\cos x=-2\sin 2x\sin x$
Substituting the values of cosine and sine expression that we have just calculated in:$\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x$
$\begin{aligned} & \left( 2\sin 2x\cos x \right)\sin x+\left( -2\sin 2x\sin x \right)\cos x \\\ & =\left( 2\sin 2x\cos x \right)\sin x-\left( 2\sin 2x\sin x \right)\cos x \\\ & =0 \\\ \end{aligned}$
As simplifying L.H.S has given 0 and which is equal to R.H.S.
Hence, we have proved L.H.S = R.H.S of the given equation.

Note: The other way of proving the given equation is:
$\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x=0$
We are going to solve the L.H.S of the above equation.
$\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x$
Multiplying $\sin x$ with the first bracket and $\cos x$ with the second bracket we get,
$\sin 3x\sin x+{{\sin }^{2}}x+\cos 3x\cos x-{{\cos }^{2}}x$
Rearranging the above equation we get,
$\sin 3x\sin x+\cos 3x\cos x-{{\cos }^{2}}x+{{\sin }^{2}}x$……..Eq. (1)
As you can see carefully, in the above equation the first two trigonometric expressions are the expansion of $\cos \left( 3x-x \right)$.
We know that $\cos \left( 3x-x \right)=\cos 3x\cos x+\sin x\sin 3x$.
Rewriting the eq. (1) as:
$\cos (3x-x)-\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$. Substituting in the above equation we get,
$\begin{aligned} & \cos 2x-\cos 2x \\\ & =0 \\\ \end{aligned}$
From the above simplification, L.H.S of the given expression has come out to be 0 which is equal to R.H.S.
Hence we have proved the L.H.S = R.H.S of the given equation.