Trigonometrical Ratios and Identities 1 Question 18
18. Let $f:(-1,1) \rightarrow R$ be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}$ for $\theta \in 0, \frac{\pi}{4} \cup \frac{\pi}{4}, \frac{\pi}{2}$. Then, the value(s) of $f \frac{1}{3}$ is/are
(a) $1-\sqrt{\frac{3}{2}}$
(b) $1+\sqrt{\frac{3}{2}}$
(c) $1-\sqrt{\frac{2}{3}}$
(d) $1+\sqrt{\frac{2}{3}}$
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Answer:
Correct Answer: 18. $A \rightarrow r ; B \rightarrow p$
Solution:
$$ \begin{array}{rlrl} f(\cos 4 \theta) & =\frac{2}{2-\sec ^{2} \theta} \\ \text { At } \quad \cos 4 \theta & =\frac{1}{3} \\ \Rightarrow \quad 2 \cos ^{2} 2 \theta-1 & =\frac{1}{3} \\ \Rightarrow \quad & \cos ^{2} 2 \theta & =\frac{2}{3} \\ \Rightarrow \quad & \cos 2 \theta & = \pm \sqrt{\frac{2}{3}} \\ \therefore & & f(\cos 4 \theta) & =\frac{2 \cdot \cos ^{2} \theta}{2 \cos ^{2} \theta-1} \\ & =\frac{1+\cos 2 \theta}{\cos 2 \theta} \\ \Rightarrow \quad f & & =1 \pm \sqrt{\frac{3}{2}} \end{array} $$
[from Eq. (ii)]
