How do you express sin(pi/ 6 ) * cos( ( 9 pi) / 8 ) without using products of trigonometric functions?

2 Answers
P dilip_k
Mar 25, 2016

=-1/2sqrt(1/2*(1+1/sqrt2))

Explanation:

sin(pi/6)*cos(pi+pi/8)= -1/2*cos(pi/8)
= -1/2sqrt(1/2*(1+cos(pi/4))) =-1/2sqrt(1/2*(1+1/sqrt2))

It may also be expressed as sum of two trigonometric functions
using formula
sinAcosB=1/2*(sin(A+B)+sin(A-B))

sin(pi/6)*cos(pi+pi/8)= -sin(pi/6)*cos(pi/8)

=-1/2*(sin(pi/6+pi/8)+sin(pi/6-pi/8))

=-1/2*(sin(7pi/24)+sin(pi/24)

Nghi N.
Mar 25, 2016

- (1/2)(cos pi/8)

Explanation:

P = sin (pi/6).cos ((9pi)/8)
Trig table --> sin (pi/6) = 1/2
cos ((9pi)/8) = cos (pi/8 + pi) = - cos (pi/8)
The product P can be expressed as:
P = - (1/2) (cos pi/8)

If being asked, we can find P by evaluating cos (pi/8), using the trig identity: cos 2a =2cos^2a - 1.

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