How do you express #cos(pi/ 4 ) * sin( ( 13 pi) / 8 ) # without using products of trigonometric functions?

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How do you express #cos(pi/ 4 ) * sin( ( 13 pi) / 8 ) # without using products of trigonometric functions?

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Answer 1 · 2016-04-12T11:42:59

#(sin((15pi)/16)+sin((11pi)/16))/2#

Explanation:

Use # sin (A+B)cos(A-B)=(sin 2A + sin 2B)/2 =#
Here, #A+B=13pi/8 and A-B=pi/4#
Add and subtract.
#2A = 15pi/8. A + 15pi/16. 2B = 11pi/8. B = 11pi/16#.
Answer: #(sin((15pi)/16)+sin((11pi)/16))/2#

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How do you express #cos(pi/ 4 ) * sin( ( 13 pi) / 8 ) # without using products of trigonometric functions?

1 Answer

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