Trigonometric Equations Question 2
Question 2 - 29 January - Shift 1
Let $f(\theta)=3(\sin ^{4}(\frac{3 \pi}{2}-\theta)+\sin ^{4}(3 \pi+\theta))-2(1-\sin ^{2} 2 \theta)$ and $S={\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}}$. If $4 \beta=\sum _{\theta \in S} \theta$, then $f(\beta)$ is equal to
(1) $\frac{11}{8}$
(2) $\frac{5}{4}$
(3) $\frac{9}{8}$
(4) $\frac{3}{2}$
Show Answer
Answer: (2)
Solution:
Formula: General solution of Trigonometric Equations, Pythagorean Identities, Periodicity Identities, Double Angle Identities, Differentiation of some elementary functions
$ \begin{aligned} & f(\theta)=3(\sin ^{4}(\frac{3 \pi}{2}-\theta)+\sin ^{4}(3 x+\theta))-2(1-\sin ^{2} 2 \theta) \\ & S={\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}} \\ & \Rightarrow f(\theta)=3(\cos ^{4} \theta+\sin ^{4} \theta)-2 \cos ^{2} 2 \theta \\ & \Rightarrow f(\theta)=3(1-\frac{1}{2} \sin ^{2} 2 \theta)-2 \cos ^{2} 2 \theta \\ & \Rightarrow f(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta \\ & =\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}(\frac{1+\cos 4 \theta}{2}) \\ & f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4} \\ & f^{\prime}(\theta)=\sin 4 \theta \\ & \Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2} \\ & \Rightarrow 4 \theta=n \pi+(-1)^{n} \frac{\pi}{3} \\ & \Rightarrow \theta=\frac{n \pi}{4}+(-1)^{n} \frac{\pi}{12} \\ & \Rightarrow \theta=\frac{\pi}{12},(\frac{\pi}{4}-\frac{\pi}{12}),(\frac{\pi}{2}+\frac{\pi}{12}),(\frac{3 \pi}{4}-\frac{\pi}{12}) \\ & \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\ & \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \end{aligned} $
