Question

Prove that $\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$.

Answer 1

In this problem, we have to prove the given trigonometric expression. We can first solve the numerator and the denominator of the LHS separately using the formulas $\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$ and $\sin x+\sin y=2\sin \dfrac{x+y}{2}\cos \dfrac{x-y}{2}$. We can substitute the values in the left-hand side part and simplify the Left-hand side part using some trigonometric identities to get the right-hand side.

Complete step by step solution:
We know that the given trigonometric expression is,
$\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$.
We can now take LHS and solve it to get the RHS.
LHS = $\dfrac{\sin 5x+\sin x-2\sin 3x}{\cos 5x-\cos x}$…….. (1)
We can solve the numerator and the denominator separately.
We now take $\sin 5x+\sin x$ and solve.
We know that,
$\sin x+\sin y=2\sin \dfrac{x+y}{2}\cos \dfrac{x-y}{2}$
We can put x = 5x and y = x in the above formula, we get

& \Rightarrow \sin 5x+\sin x=2\sin \dfrac{5x+x}{2}\cos \dfrac{5x-x}{2} \\\ & \Rightarrow \sin 5x+\sin x=2\sin \dfrac{6x}{2}\cos \dfrac{4x}{2} \\\ & \Rightarrow \sin 5x+\sin x=2\sin 3x\cos 2x.......\left( 2 \right) \\\ \end{aligned}$$ We now take $$\cos 5x-\cos x$$ and solve. We know that, $$\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$$ We can put x = 5x and y = x in the above formula, we get $$\begin{aligned} & \Rightarrow \cos 5x-\cos x=-2\sin \dfrac{5x+x}{2}\sin \dfrac{5x-x}{2} \\\ & \Rightarrow \cos 5x-\cos x=-2\sin \dfrac{6x}{2}\sin \dfrac{4x}{2} \\\ & \Rightarrow \cos 5x-\cos x=-2\sin 3x\sin 2x.......\left( 3 \right) \\\ \end{aligned}$$ We can now substitute (2) and (3) in (1), we get LHS = $$\dfrac{2\sin 3x\cos 2x-2\sin 3x}{-2\sin 3x\sin 2x}$$ We can now take the common term$$2\sin 3x$$ separately, in the numerator, we get LHS = $$\dfrac{2\sin 3x\left( \cos 2x-1 \right)}{-2\sin 3x\sin 2x}$$ We can now cancel the similar terms in the numerator and the denominator and we can take the negative sign to numerator, we get LHS = $$\dfrac{1-\cos 2x}{\sin 2x}$$……. (4) We can now use a trigonometric identity, $$\begin{aligned} & \cos 2x=1-2{{\sin }^{2}}x \\\ & \sin 2x=2\sin x\cos x \\\ \end{aligned}$$ We can apply the above identities in (4), we get LHS = $$\dfrac{1-\left( 1-2{{\sin }^{2}}x \right)}{2\cos x\sin x}$$ We can now simplify the above step, we get LHS = $$\dfrac{1-1+2{{\sin }^{2}}x}{2\sin x\cos x}=\dfrac{2{{\sin }^{2}}x}{2\sin x\cos x}$$ We can now cancel the similar terms in the above step, we get LHS = $$\dfrac{\sin x}{\cos x}$$ LHS = $$\tan x\text{ }\because \dfrac{\sin x}{\cos x}=\tan x$$ LHS = RHS. Hence proved that $$\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$$. **Note:** Students make mistakes, while using correct trigonometric formulas as we will make mistakes in writing sine and cosine parts in the formula. We should always remember the exact formula to prove the given expressions. We should also remember some identities like $$\cos 2x=1-2{{\sin }^{2}}x$$ and $$\sin 2x=2\sin x\cos x$$ to be used in these types of problems.