How do you express #cos(pi/ 3 ) * sin( ( 5 pi) / 8 ) # without using products of trigonometric functions?

2 Answers
Aug 1, 2016

It may be "cheating", but I would just substitute #1/2# for #cos(\pi/3)#.

Explanation:

You're probably supposed to use the identity

#cos a sin b = (1/2)(sin(a+b)-sin(a-b))#.

Put in #a=\pi/3={8\pi}/24, b={5\pi}/8={15\pi}/24#.

Then

#cos(\pi/3)sin({5*pi}/8)=(1/2)(sin({23*\pi}/24)-sin({-7*pi}/24))#

#=(1/2)(sin({\pi}/24)+sin({7*\pi}/24))#

where in the last line we use #sin(\pi-x)=sin(x)# and #sin(-x)=-sin(x)#.

As you can see, this is unwieldy compared with just putting in #cos(pi/3)=1/2#. The trigonometric product-sum and product-difference relations are more useful when you can't evaluate either factor in the product.

Aug 2, 2016

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

Explanation:

#P = cos (pi/3).sin ((5pi)/8)#
Trig table --> #cos (pi/3) = 1/2#
Trig unit circle and property of complementary arcs -->
#sin ((5pi)/8) = sin (pi/8 + (4pi)/8) = sin (pi/8 + pi/2) = #
#= - cos (pi/8).#
P can be expressed as:
#P = - (1/2)cos (pi/8)#
NOTE. We can evaluate #cos (pi/8)# by using the trig identity:
#1 + cos (pi/4) = 2cos^2 (pi/8)#