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Sin pi/4sinpi/6 = 1/2[__pi/12-cos5pi/12]

Sin pi/4sinpi/6 = 1/2[__pi/12-cos5pi/12]
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Sin pi/4sinpi/6 = 1/2[__pi/12-cos5pi/12]

User Kelvin Tan
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1 Answer

4 votes

Answer:

Explanation:

2 sin a sin b=cos(a-b)-cos (a+b)


2 sin (\pi )/(4) sin (\pi )/(6)=cos ((\pi )/(4)-(\pi )/(6))-cos ((\pi )/(4) +(\pi )/(6) )\\=cos ((3 \pi-2 \pi )/(12) )-cos ((3 \pi+2\pi)/(12) )\\=cos ((\pi )/(12) )-cos( (5 \pi)/(12) )\\sin ( \pi)/(4) sin\frac {\pi}{6}=(1)/(2)[cos (\pi)/(12) -cos (5 \pi)/(12) ]

User Merr Leader
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