Question

What is the value of $[(\cos3\theta + 2\cos5\theta + \cos7\theta) \div (\cos\theta + 2\cos3\theta + \cos5\theta)] + \sin2\theta.\tan3\theta$?

• A. $\cos2\theta$
• B. $\sin2\theta$
• C. $\tan2\theta$
• D. $\cot\theta.\sin2\theta$

Answer 1

Answer: (A)

$\cos2\theta$

Answer 2

The problem requires simplifying a trigonometric expression using various identities.

The given expression is: $E = \left[\frac{\cos3\theta + 2\cos5\theta + \cos7\theta}{\cos\theta + 2\cos3\theta + \cos5\theta}\right] + \sin2\theta.\tan3\theta$

Let's simplify the numerator of the first term, $N = \cos3\theta + 2\cos5\theta + \cos7\theta$. We can rearrange the terms and use the sum-to-product identity $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$: $N = (\cos3\theta + \cos7\theta) + 2\cos5\theta$ $N = 2\cos\left(\frac{3\theta+7\theta}{2}\right)\cos\left(\frac{7\theta-3\theta}{2}\right) + 2\cos5\theta$ $N = 2\cos5\theta\cos2\theta + 2\cos5\theta$ Factor out $2\cos5\theta$: $N = 2\cos5\theta(\cos2\theta + 1)$ Now, use the identity $1+\cos2\theta = 2\cos^2\theta$: $N = 2\cos5\theta(2\cos^2\theta)$ $N = 4\cos5\theta\cos^2\theta$

Next, let's simplify the denominator of the first term, $D = \cos\theta + 2\cos3\theta + \cos5\theta$. Rearrange and use the sum-to-product identity: $D = (\cos\theta + \cos5\theta) + 2\cos3\theta$ $D = 2\cos\left(\frac{\theta+5\theta}{2}\right)\cos\left(\frac{5\theta-\theta}{2}\right) + 2\cos3\theta$ $D = 2\cos3\theta\cos2\theta + 2\cos3\theta$ Factor out $2\cos3\theta$: $D = 2\cos3\theta(\cos2\theta + 1)$ Use the identity $1+\cos2\theta = 2\cos^2\theta$: $D = 2\cos3\theta(2\cos^2\theta)$ $D = 4\cos3\theta\cos^2\theta$

Now, substitute the simplified numerator and denominator back into the first term of the expression: $\frac{N}{D} = \frac{4\cos5\theta\cos^2\theta}{4\cos3\theta\cos^2\theta}$ Assuming $\cos^2\theta \neq 0$, we can cancel $\cos^2\theta$: $\frac{N}{D} = \frac{\cos5\theta}{\cos3\theta}$

Substitute this back into the original expression: $E = \frac{\cos5\theta}{\cos3\theta} + \sin2\theta.\tan3\theta$ Rewrite $\tan3\theta$ as $\frac{\sin3\theta}{\cos3\theta}$: $E = \frac{\cos5\theta}{\cos3\theta} + \sin2\theta \left(\frac{\sin3\theta}{\cos3\theta}\right)$ Combine the terms over a common denominator $\cos3\theta$: $E = \frac{\cos5\theta + \sin2\theta\sin3\theta}{\cos3\theta}$

Now, use the cosine addition formula $\cos(A+B) = \cos A\cos B - \sin A\sin B$. Let $A=2\theta$ and $B=3\theta$, then $\cos(2\theta+3\theta) = \cos5\theta$: $\cos5\theta = \cos2\theta\cos3\theta - \sin2\theta\sin3\theta$ Substitute this expression for $\cos5\theta$ into the numerator of $E$: Numerator $= (\cos2\theta\cos3\theta - \sin2\theta\sin3\theta) + \sin2\theta\sin3\theta$ Numerator $= \cos2\theta\cos3\theta$

So, the expression $E$ becomes: $E = \frac{\cos2\theta\cos3\theta}{\cos3\theta}$ Assuming $\cos3\theta \neq 0$, we can cancel $\cos3\theta$: $E = \cos2\theta$

The value of the expression is $\cos2\theta$.