If + = 12, then 16(2+4+6) is equal to:

Question

If $\sin	heta+\cos	heta = \dfrac12$, then $16(\sin2	heta+\cos4	heta+\sin6	heta)$ is equal to:

Accepted Answer

Squaring: $1+\sin2\theta = \tfrac14 \implies \sin2\theta = -\tfrac34$.

$$\cos4\theta = 1-2\sin^22\theta = 1-2\left(\tfrac9{16}\right) = -\tfrac18$$

$$\sin6\theta = 3\sin2\theta-4\sin^32\theta = (3-4\sin^22\theta)\sin2\theta = \left(3-\tfrac94\right)\left(-\tfrac34\right) = \tfrac34\cdot\left(-\tfrac34\right) = -\tfrac9{16}$$

$$16\left[-\tfrac34-\tfrac18-\tfrac9{16}\right] = 16\left[-\tfrac{12}{16}-\tfrac2{16}-\tfrac9{16}\right] = 16\times\left(-\tfrac{23}{16}\right) = -23$$

Solutions · 1