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

Answer: The proof is complete, as $\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$ .

Step-by-step solution

Step 1: Rearrange the numerator

We rearrange the terms in the numerator to group $\sin 5x$ and $\sin x$ together. This allows us to apply the sum-to-product formula for sine.

Step 2: Apply sum-to-product formulas

We will use the sum-to-product formula for sine in the numerator and the difference-to-product formula for cosine in the denominator. These identities help simplify trigonometric expressions.

Step 3: Simplify the numerator

Applying the sum-to-product formula to $(\sin 5x + \sin x)$ gives $2 \sin 3x \cos 2x$ . Then, we factor out $2 \sin 3x$ from the expression.

Step 4: Simplify the denominator

Applying the difference-to-product formula to $(\cos 5x - \cos x)$ gives $-2 \sin 3x \sin 2x$ .

Step 5: Substitute and simplify

Substitute the simplified numerator and denominator back into the original expression. We can cancel out $2 \sin 3x$ from both the numerator and the denominator.

Step 6: Use double angle identities

We use the double angle identities for $\cos 2x$ and $\sin 2x$ to further simplify the expression. Specifically, $\cos 2x - 1 = -2 \sin^2 x$ .

Step 7: Final simplification

Substitute the double angle identities into the expression. Cancel out $-2 \sin x$ from the numerator and denominator. The remaining term $\frac{\sin x}{\cos x}$ is equal to $\tan x$ .

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Prove that =sin⁡5x−2sin⁡3x+sin⁡xcos⁡5x−cos⁡x=tan⁡x=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x=cos5x−cosxsin5x−2sin3x+sinx​=tanx