Q62 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

If $P+Q+R=\frac{\pi}{4}$, then $\cos \left(\frac{\pi}{8}-P\right)+\cos \left(\frac{\pi}{8}-Q\right)+\cos \left(\frac{\pi}{8}-R\right)=$

  • Correct answer
    A
    $4 \cos \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2}-\cos \frac{\pi}{8}$
  • B
    $4 \cos \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2}+\cos \frac{\pi}{8}$
  • C
    $4 \sin \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2}-\cos \frac{\pi}{8}$
  • D
    $4 \sin \frac{P}{2} \cos \frac{Q}{2} \cos \frac{R}{2}+\cos \frac{\pi}{8}$

Solution

$\begin{aligned} & \text { Given that } \mathrm{P}+\mathrm{Q}+\mathrm{R}=\frac{\pi}{4} \\ & \begin{array}{l}\therefore \cos \left(\frac{\pi}{8}-\mathrm{P}\right)+\cos \left(\frac{\pi}{8}-\mathrm{Q}\right)+\cos \left(\frac{\pi}{8}-\mathrm{R}\right) \\ =2 \cos \frac{1}{2}\left(\frac{\pi}{4}-(\mathrm{P}+\mathrm{Q})\right) \cos \frac{1}{2}(\mathrm{Q}-\mathrm{P})+ \\ \qquad \cos \left(\frac{\pi}{8}-\mathrm{R}\right)+\cos \frac{\pi}{8}-\cos \frac{\pi}{8} \\ =2 \cos \frac{\mathrm{R}}{2} \cdot \cos \frac{1}{2}(\mathrm{Q}-\mathrm{P})+2 \cos \frac{1}{2}\left(\frac{\pi}{4}-\mathrm{R}\right) \cdot \cos \frac{\mathrm{R}}{2}-\cos \frac{\pi}{8} \\ =2 \cos \frac{\mathrm{R}}{2} \cdot \cos \frac{1}{2}(\mathrm{Q}-\mathrm{P})+2 \cos \frac{1}{2}(\mathrm{P}+\mathrm{Q}) \cdot \cos \frac{\mathrm{R}}{2}-\cos \frac{\pi}{8} \\ =2 \cos \frac{\mathrm{R}}{2}\left[\cos \frac{1}{2}(\mathrm{P}-\mathrm{Q})+\cos \frac{1}{2}(\mathrm{P}+\mathrm{Q})\right]-\cos \frac{\pi}{8} \\ =4 \cos \frac{\mathrm{R}}{2} \cdot \cos \frac{\mathrm{P}}{2} \cdot \cos \frac{\mathrm{Q}}{2}-\cos \frac{\pi}{8}\end{array}\end{aligned}$

Hence, the correct answer is option: 'A'.

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