How do you simplify #sin(pi/8)cos(pi/8)# using double-angle identities?

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Answer 1 · 2015-04-21T19:16:43

#sin(2A) = 2sinAcosA#

#cos(2A) = cos^2A - sin^2A#
#color(white)"ssssssss"# #=1-2sin^2A#
#color(white)"ssssssss"# #=2cos^2A - 1#

#tan(2A) = (2tanA)/(1-tan^2A)#

Which one or ones look post promising for rewriting

#sin(pi/8)cos(pi/8)#

If you said anything other than the first one, you're incorrect.

From #sin(2A) = 2sinAcosA#, we may conclude that:

#sinAcosA = 1/2 sin(2A)#, so the answer is

#sin(pi/8)cos(pi/8) = 1/2 sin(2 pi/8) = 1/2 sin(pi/4)=1/2 sqrt2/2=sqrt2/4#