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

2 Answers
Nghi N.
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)

Shwetank Mauria
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)

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