How do you express sin(pi/ 6 ) * cos( ( 15 pi) / 12 ) without using products of trigonometric functions?

2 Answers
Nghi N.
Mar 26, 2016

-sqrt2/4

Explanation:

P = sin (pi/6)cos ((15pi)/12).
Trig table --> sin pi/6 = 1/2
cos ((15pi)/12) = cos ((-9pi)/12 + 2pi) = cos ((-3pi)/4) =
= cos ((3pi)/4) = -sqrt2/2.
P = (1/2)(-sqrt2/2) = -sqrt2/4

Bdub
Apr 4, 2016

sin (pi/6) cos ((15pi)/12)=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)

Explanation:

2 sin A cos B=sin (A+B) + sin (A-B)
sin A cos B=1/2 (sin (A+B) + sin (A-B))
A=pi/6, B=(15pi)/12
sin (pi/6) cos ((15pi)/12)=1/2 (sin (pi/6+(15pi)/12) + sin (pi/6-(15pi)/12))
=1/2 (sin ((17pi)/12) + sin (-(13pi)/12))
=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)

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