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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