MHT CET · Maths · Trigonometric Ratios & Identities
If \(\cos 2 \theta=\sin \propto, \quad\) then \(\theta=\)
- A \(2 n \pi \pm\left(\frac{\pi}{2}-\alpha\right), n \in z\)
- B \(n \pi \pm\left(\frac{\pi}{4}+\frac{\alpha}{2}\right), n \in z\)
- C \(\frac{1}{2}\left[n \pi+(-1)^{n} \propto\right], n \in z\)
- D \(n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), n \in z\)
Answer & Solution
Correct Answer
(D) \(n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), n \in z\)
Step-by-step Solution
Detailed explanation
We have \(\cos 2 \theta=\sin \alpha \Rightarrow \cos 2 \theta=\cos (90-\alpha)\) When \(\cos \theta=\cos \alpha\), we get \(\theta=2 n \pi \pm \alpha, n \in Z\) \(\therefore \quad 2 \theta=2 \mathrm{n} \pi \pm(90-\alpha)\)
\(\theta=\frac{2 n \pi}{2} \pm\left(\frac{90-\alpha}{2}\right) \Rightarrow \theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\)
\(\theta=\frac{2 n \pi}{2} \pm\left(\frac{90-\alpha}{2}\right) \Rightarrow \theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\)
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