MCQ
$8\sin\frac{\text{x}}{8}\cos\frac{\text{x}}{2}\cos\frac{\text{x}}{4}\cos\frac{\text{x}}{8}$ is equal to:
- A$8\cos\text{x}$
- B$\cos\text{x}$
- C$8\sin\text{x}$
- D$\sin\text{x}$
Solution:
we have,
$=8\sin\text{x}8\cos\text{x}2\cos\text{x}4\cos\text{x}8$
$=4\times2\sin\text{x}8\cos\text{x}2\cos\text{x}2$
$=4\times\sin\text{x}\cos\text{x}2\cos\text{x}4$
$=2\times2\sin\text{x}4\cos\text{x}4\cos\text{x}2$
$=2\times\sin2\cos\text{x}2$
$=\sin\text{x}$
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