$= \sin \frac{\pi}{14} \cos (\frac{\pi}{2} -\frac{3\pi}{14}) \cos(\frac{\pi}{2} -\frac{5\pi}{14})$
$= \sin \frac{\pi}{14} \cos (\frac{4\pi}{14}) \cos (\frac{2 \pi}{14})$
$= \sin \frac{\pi}{14} \cos (\frac{2\pi}{14}) \cos (\frac{4 \pi}{14})$$= \dfrac{\cos \frac{\pi}{14} \sin \frac{\pi}{14} \cos (\frac{2\pi}{14}) \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{2}\frac{\sin \frac{2\pi}{14} \cos (\frac{2\pi}{14}) \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{4}\dfrac{\sin \frac{4\pi}{14} \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin \frac{8\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\pi - \frac{8\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\frac{6\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} - \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} - \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\frac{\cos \frac{\pi}{14}}{\cos \frac{\pi}{14}} $
$=\frac{1}{8}$
edited the above based on the comment to keep the flow
5 comments:
Is there a problem in typing,in 4the line that used sin instead of cos.
the answer is correct.
I spent a long time but could not solve this from the original Loney book. Nice and clean solution ! good job.
im 15 and my teacher give me to solve 124knuifas problems like that, but when i se ,,pi"-omfg,im dead.
Good solution
Good Job Well Done......
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