$\frac{\cos 3x + \cos x}{\sin 3x - \sin x}$
Simplify Trigonometric Expression
The question asks for the value of the trigonometric expression $\frac{\cos 3x + \cos x}{\sin 3x - \sin x}$. We will use sum-to-product and difference-to-product trigonometric identities to simplify this expression.
Apply Trigonometric Identities
Use the sum-to-product identity for the numerator:
$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$
For the numerator, $A=3x$ and $B=x$:
$\cos 3x + \cos x = 2 \cos\left(\frac{3x+x}{2}\right) \cos\left(\frac{3x-x}{2}\right) = 2 \cos\left(\frac{4x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \cos(2x) \cos(x)$
Use the difference-to-product identity for the denominator:
$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$
For the denominator, $A=3x$ and $B=x$:
$\sin 3x - \sin x = 2 \cos\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = 2 \cos\left(\frac{4x}{2}\right) \sin\left(\frac{2x}{2}\right) = 2 \cos(2x) \sin(x)$
Simplify the Expression
Now, substitute the simplified numerator and denominator back into the original expression:
$\frac{\cos 3x + \cos x}{\sin 3x - \sin x} = \frac{2 \cos(2x) \cos(x)}{2 \cos(2x) \sin(x)}$
Cancel out the common term $2 \cos(2x)$ (assuming $\cos(2x) \neq 0$):
$\frac{\cos(x)}{\sin(x)}$
Final Result
The expression simplifies to $\cot(x)$.
Final Answer: The final answer is $\boxed{\cot x}$
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Important Questions from Trigonometric Ratios and Identities
What is the ratio of the greatest to the smallest value of 2 – 2 sin x – sin 2x, 0 ≤ x ≤ (π/2)?
If sinθ = \(\frac{4}{5}\) , Find the value of sin3θ
If x, y are acute angles, where 0 < x + y < 90° and sin(3x - 40°) = cos (3y + 40°), then the value of tan (x + y) is equal to
What is (1 + cot θ - cosec θ)(1 + tan θ + sec θ) equal to?
If \(\sin \theta =\frac{3}{5}\) and \(\cos \theta =\frac{4}{5}\) , then the value of \(\frac{1+\tan \theta}{1-\cot \theta}\) is:
