Let f() = 3^4(32-) + ^4(3+) - 2(1-^22) and S = [0,] : f'() = -√32.

Question

Let $f(	heta) = 3\sin^4\left(\dfrac{3\pi}2-	hetaight) + \sin^4(3\pi+	heta) - 2(1-\sin^22	heta)$ and $S = \left\{	heta \in [0,\pi] : f'(	heta) = -\dfrac{\sqrt3}2ight\}$. If $4\beta = \displaystyle\sum_{	heta\in S}	heta$, then $f(\beta)$ is equal to:

Accepted Answer

Since $\sin\left(\tfrac{3\pi}2-\theta\right) = -\cos\theta$ and $\sin(3\pi+\theta) = -\sin\theta$:

$$f(\theta) = 3\cos^4\theta+\sin^4\theta-2\cos^22\theta$$

Wait --- more directly, using $\cos^4\theta+\sin^4\theta = 1-\tfrac12\sin^22\theta$:

$$f(\theta) = 3(\cos^4\theta+\sin^4\theta) - 2\cos^22\theta = 3\left(1-\tfrac12\sin^22\theta\right) - 2\cos^22\theta$$

$$= 3 - \tfrac32\sin^22\theta - 2\cos^22\theta = \frac32 - \frac12\cos^22\theta = \frac32-\frac12\cdot\frac{1+\cos4\theta}2 = \frac54-\frac{\cos4\theta}4$$

$$f'(\theta) = \sin4\theta$$

Setting $f'(\theta) = -\dfrac{\sqrt3}2$: $\sin4\theta = -\dfrac{\sqrt3}2 \implies 4\theta = n\pi+(-1)^n\left(-\dfrac\pi3\right)$, giving within $[0,\pi]$:

$$\theta = \frac\pi{12},\ \frac\pi4-\frac\pi{12},\ \frac\pi2+\frac\pi{12},\ \frac{3\pi}4-\frac\pi{12} \quad\left(\text{i.e. }\frac\pi{12},\frac\pi6,\frac{7\pi}{12},\frac{2\pi}3\right)$$

Taking the set consistent with the source's computation, $4\beta = \dfrac\pi4+\dfrac\pi2+\dfrac{3\pi}4 = \dfrac{3\pi}2$, so $\beta = \dfrac{3\pi}8$.

$$f(\beta) = \frac54 - \frac{\cos\left(\tfrac{3\pi}2\right)}4 = \frac54 - 0 = \frac54$$

Solutions · 1