Question

For any $\theta \in (\frac{\pi}{4},\frac{\pi}{2})$, the expression 3(sin$\theta$ - cos$\theta$)$^4$ + 6(sin$\theta$ + cos$\theta$)$^2$ + 4sin^6$$\theta equals :

[JEE-MAIN 2019]

• A. 13 - 4 cos6$\theta$
• B. 13 - 4 cos4$\theta$ + 2 sin^2$$\thetacos2$\theta$
• C. 13 - 4 cos2$\theta$ + 6 cos4$\theta$
• D. 13 - 4 cos2$\theta$ + 6 sin2$\theta$cos2$\theta$

Answer 1

No option matches

Answer 2

Let the given expression be $E$.

$E = 3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta$

We know the following trigonometric identities:

1. $(\sin\theta - \cos\theta)^2 = \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = 1 - \sin(2\theta)$

2. $(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + \sin(2\theta)$

3. $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$

4. $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$

5. $\cos(3A) = 4\cos^3A - 3\cos A \implies \cos^3A = \frac{\cos(3A) + 3\cos A}{4}$

Let's simplify the first two terms:

$3(\sin\theta - \cos\theta)^4 = 3[(\sin\theta - \cos\theta)^2]^2 = 3(1 - \sin(2\theta))^2$

$6(\sin\theta + \cos\theta)^2 = 6(1 + \sin(2\theta))$

Combining these two parts:

$3(1 - \sin(2\theta))^2 + 6(1 + \sin(2\theta))$

$= 3(1 - 2\sin(2\theta) + \sin^2(2\theta)) + 6 + 6\sin(2\theta)$

$= 3 - 6\sin(2\theta) + 3\sin^2(2\theta) + 6 + 6\sin(2\theta)$

$= 9 + 3\sin^2(2\theta)$

Now, let's simplify the third term, $4\sin^6\theta$:

$4\sin^6\theta = 4(\sin^2\theta)^3$

Substitute $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$:

$4\left(\frac{1 - \cos(2\theta)}{2}\right)^3 = 4\frac{(1 - \cos(2\theta))^3}{8} = \frac{1}{2}(1 - \cos(2\theta))^3$

Expand $(1 - \cos(2\theta))^3$ using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$\frac{1}{2}(1 - 3\cos(2\theta) + 3\cos^2(2\theta) - \cos^3(2\theta))$

Now, substitute $\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}$ and $\cos^3(2\theta) = \frac{\cos(6\theta) + 3\cos(2\theta)}{4}$:

$\frac{1}{2}\left(1 - 3\cos(2\theta) + 3\left(\frac{1 + \cos(4\theta)}{2}\right) - \left(\frac{\cos(6\theta) + 3\cos(2\theta)}{4}\right)\right)$

$= \frac{1}{2}\left(1 - 3\cos(2\theta) + \frac{3}{2} + \frac{3}{2}\cos(4\theta) - \frac{1}{4}\cos(6\theta) - \frac{3}{4}\cos(2\theta)\right)$

Combine constant terms and $\cos(2\theta)$ terms:

$= \frac{1}{2}\left(\left(1 + \frac{3}{2}\right) - \left(3 + \frac{3}{4}\right)\cos(2\theta) + \frac{3}{2}\cos(4\theta) - \frac{1}{4}\cos(6\theta)\right)$

$= \frac{1}{2}\left(\frac{5}{2} - \frac{15}{4}\cos(2\theta) + \frac{3}{2}\cos(4\theta) - \frac{1}{4}\cos(6\theta)\right)$

$= \frac{5}{4} - \frac{15}{8}\cos(2\theta) + \frac{3}{4}\cos(4\theta) - \frac{1}{8}\cos(6\theta)$

Now, add this to $9 + 3\sin^2(2\theta)$.

First, express $3\sin^2(2\theta)$ in terms of $\cos(4\theta)$:

$3\sin^2(2\theta) = 3\left(\frac{1 - \cos(4\theta)}{2}\right) = \frac{3}{2} - \frac{3}{2}\cos(4\theta)$.

So, the total expression is:

$E = 9 + \left(\frac{3}{2} - \frac{3}{2}\cos(4\theta)\right) + \left(\frac{5}{4} - \frac{15}{8}\cos(2\theta) + \frac{3}{4}\cos(4\theta) - \frac{1}{8}\cos(6\theta)\right)$

Combine constant terms: $9 + \frac{3}{2} + \frac{5}{4} = \frac{36+6+5}{4} = \frac{47}{4}$.

Combine $\cos(2\theta)$ terms: $-\frac{15}{8}\cos(2\theta)$.

Combine $\cos(4\theta)$ terms: $-\frac{3}{2}\cos(4\theta) + \frac{3}{4}\cos(4\theta) = \left(-\frac{6}{4} + \frac{3}{4}\right)\cos(4\theta) = -\frac{3}{4}\cos(4\theta)$.

Combine $\cos(6\theta)$ terms: $-\frac{1}{8}\cos(6\theta)$.

So, $E = \frac{47}{4} - \frac{15}{8}\cos(2\theta) - \frac{3}{4}\cos(4\theta) - \frac{1}{8}\cos(6\theta)$.

This expression does not directly match any of the options. This suggests there might be a simpler way or an identity that was missed. The problem is likely flawed.