How do you express #sin(pi/ 8 ) * cos(( ( 5 pi) / 4 ) # without using products of trigonometric functions?

2 Answers
May 21, 2016

#- (sqrt2/2)sin (pi/8)#

Explanation:

#P = sin (pi/8).cos ((5pi)/4)#
Trig table and property of supplementary arcs -->
#cos ((5pi)/4) = cos (pi/4 + pi) = - cos (pi/4) = -sqrt2/2.#
Therefor, P can be expressed as
#P = - (sqrt2/2)sin (pi/8)#

Note. We can evaluate P by finding exact value of sin (pi/8), using the trig identity:
#cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)#

May 21, 2016

#sin(pi/8)cos((5pi)/4)=-1/4sqrt(4-2sqrt2)#

Explanation:

#sin(pi/8)cos((5pi)/4)#

Let us first calculate

#cos((5pi)/4)=cos(pi/4+pi)=-cos(pi/4)=-1/sqrt2#

For #sin(pi/8)#, let us use the identity #cos2theta=1-2sin^2theta#

Hence, #cos(pi/4)=1-2sin^2(pi/8)# or

#1/sqrt2=1-2sin^2(pi/8)# or

#2sin^2(pi/8)=1-1/sqrt2=1-sqrt2/2=(2-sqrt2)/2#

or #sin^2(pi/8)=(2-sqrt2)/4#

or #sin(pi/8)=1/2sqrt(2-sqrt2)#

Hence, #sin(pi/8)cos((5pi)/4)=1/2sqrt(2-sqrt2)xx-sqrt2/2#

or #sin(pi/8)cos((5pi)/4)=-1/4sqrt(4-2sqrt2)#