If P+Q+R=\frac{\pi}{4}, then \cos (\frac{\pi}{8}-P) +\cos (\frac{\pi}{8}-Q) +\cos (\frac{\pi}{8}-R) = [AP EAPCET 21 May 2024 Shift 1]

Correct Answer is : 4cosP2cosQ2cosR2−cosπ8

4 \cos \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2}-\cos \frac{\pi}{8}

4 \cos \frac{P}{2} \cos \frac{Q}{2} \sin \frac{R}{2}+\cos \frac{\pi}{8}

4 \sin \frac{P}{2} \sin \frac{Q}{2} \sin \frac{R}{2}-\cos \frac{\pi}{8}

4 \sin \frac{P}{2} \cos \frac{Q}{2} \sin \frac{R}{2}+\cos \frac{\pi}{8}

Trigonometric Ratios and Identities

Section: Mathematics

Question : 38 of 50

Marks: +1, -0

If P+Q+R=‌

, then cos(‌

−P)+cos(‌

−Q)+cos(‌

−R)=

[AP EAPCET 21 May 2024 Shift 1]

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