Let S = [0,2] : 8^2^2 + 8^2^2 = 16.

Question

Let $S = \left\{	heta \in [0,2\pi] : 8^{2\sin^2	heta} + 8^{2\cos^2	heta} = 16ight\}$. Then $n(S) + \displaystyle\sum_{	heta\in S}\sec\left(\dfrac\pi4+2	hetaight)\csc\left(\dfrac\pi4+2	hetaight)$ is equal to:

Accepted Answer

By AM--GM on $8^{2\sin^2\theta}$ and $8^{2\cos^2\theta}$:

$$\frac{8^{2\sin^2\theta}+8^{2\cos^2\theta}}2 \ge 8^{\sin^2\theta+\cos^2\theta} = 8 \implies 8^{2\sin^2\theta}+8^{2\cos^2\theta}\ge16$$

with equality iff $8^{2\sin^2\theta}=8^{2\cos^2\theta}$, i.e. $\sin2\theta=\cos2\theta$. So $\theta = \dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4$, giving $n(S)=4$.

$$\sec\left(\tfrac\pi4+2\theta\right)\csc\left(\tfrac\pi4+2\theta\right) = \frac2{\sin\left(\tfrac\pi2+4\theta\right)} = 2\csc\left(\tfrac\pi2+4\theta\right)$$

Evaluating at the four values of $\theta$, each term gives $2\csc\left(\tfrac\pi2+n\pi\right) = -2$, so the sum of the four terms is $-8$.

$$n(S) + \sum(\cdots) = 4 + (-8) = -4$$

Solutions · 1