Question

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

Answer 1

`- (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)`

Answer 2

`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)`