Question

If $\sin\theta + \cos\theta = \frac{1}{2}$, then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to :

• A. 23
• B. -23
• C. 27
• D. -27

Hint:

When given $\sin\theta \pm \cos\theta = k$, the first step is almost always to square the equation to get a value for $\sin(2\theta)$. From there, you can use double and triple angle formulas to find higher multiples of $\theta$.

Answer 1

The Correct Option is B

Given $\sin\theta + \cos\theta = \frac{1}{2}$.
Square both sides: $(\sin\theta + \cos\theta)^2 = (\frac{1}{2})^2$.
$\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = \frac{1}{4}$.
$1 + \sin(2\theta) = \frac{1}{4}$.
$\sin(2\theta) = \frac{1}{4} - 1 = -\frac{3}{4}$.
Now let's find the other terms in the expression.
We need $\cos(4\theta)$. We can use the identity $\cos(2A) = 1 - 2\sin^2(A)$.
$\cos(4\theta) = \cos(2(2\theta)) = 1 - 2\sin^2(2\theta)$.
$\cos(4\theta) = 1 - 2\left(-\frac{3}{4}\right)^2 = 1 - 2\left(\frac{9}{16}\right) = 1 - \frac{18}{16} = 1 - \frac{9}{8} = -\frac{1}{8}$.
Next, we need $\sin(6\theta)$. We use the identity $\sin(3A) = 3\sin A - 4\sin^3 A$.
$\sin(6\theta) = \sin(3(2\theta)) = 3\sin(2\theta) - 4\sin^3(2\theta)$.
Substitute the value of $\sin(2\theta) = -3/4$:
$\sin(6\theta) = 3\left(-\frac{3}{4}\right) - 4\left(-\frac{3}{4}\right)^3$.
$= -\frac{9}{4} - 4\left(-\frac{27}{64}\right) = -\frac{9}{4} + \frac{27}{16}$.
$= \frac{-36+27}{16} = -\frac{9}{16}$.
Now, substitute all these values back into the expression we need to calculate.
Expression = $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$.
$= 16\left(-\frac{3}{4} - \frac{1}{8} - \frac{9}{16}\right)$.
Find a common denominator, which is 16.
$= 16\left(\frac{-12 - 2 - 9}{16}\right)$.
$= 16\left(\frac{-23}{16}\right) = -23$.