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(sin 6x-sin 2x)/(cos 6x-cos 2x)=-cot 4x

$\frac{\sin 6 x-\sin 2 x}{\cos 6 x-\cos 2 x}=-\cot 4 x$

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Simplify variable expressions using properties

Full solution

Q.

\frac{\sin 6 x-\sin 2 x}{\cos 6 x-\cos 2 x}=-\cot 4 x

Apply Formulas: Use the sine and cosine subtraction formulas: $\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$ and $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ .
Calculate $\sin(4x)$ and $\cos(4x)$ : Apply the formulas to $\sin(6x - 2x)$ and $\cos(6x - 2x)$ to get $\sin(4x)$ and $\cos(4x)$ respectively.
Substitute in Original Expression: Substitute $\sin(4x)$ for $\sin(6x - 2x)$ and $\cos(4x)$ for $\cos(6x - 2x)$ in the original expression: $(\sin 4x) / (\cos 4x)$ .
Recognize Definition: Recognize that $\frac{\sin 4x}{\cos 4x}$ is the definition of $\tan(4x)$ .
Recall Cotangent Relationship: Recall that $\cot(x) = \frac{1}{\tan(x)}$ , so $\cot(4x) = \frac{1}{\tan(4x)}$ .
Final Simplification: Therefore, $(\sin 4x) / (\cos 4x) = \tan(4x)$ , which means the original expression simplifies to $-\tan(4x)$ , not $-\cot(4x)$ as suggested.

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