TS EAMCET · Maths · Trigonometric Equations
If \(A+B+C=\frac{\pi}{4}\), then \(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}-\cos \frac{\pi}{8}=\)
- A \(\cos \left(\frac{\pi}{4}-A\right)+\cos \left(\frac{\pi}{4}-B\right)+\cos \left(\frac{\pi}{4}-C\right)\)
- B \(\cos \left(\frac{\pi}{8}-A\right)+\cos \left(\frac{\pi}{8}-B\right)+\cos \left(\frac{\pi}{8}-C\right)\)
- C \(\sin \left(\frac{\pi}{4}-A\right)+\sin \left(\frac{\pi}{4}-B\right)+\sin \left(\frac{\pi}{4}-C\right)\)
- D \(\sin \left(\frac{\pi}{8}-A\right)+\sin \left(\frac{\pi}{8}-B\right)+\sin \left(\frac{\pi}{8}-C\right)\)
Answer & Solution
Correct Answer
(C) \(\sin \left(\frac{\pi}{4}-A\right)+\sin \left(\frac{\pi}{4}-B\right)+\sin \left(\frac{\pi}{4}-C\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } 2\left(2 \cos \frac{A}{2} \cos \frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8} \\ & =2\left[\cos \left(\frac{A}{2}+\frac{B}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2} \\ & =2\left[\cos…
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