How do you express cos(π3)sin(5π8) without using products of trigonometric functions?

2 Answers
Aug 1, 2016

It may be "cheating", but I would just substitute 12 for cos(π3).

Explanation:

You're probably supposed to use the identity

cosasinb=(12)(sin(a+b)sin(ab)).

Put in a=π3=8π24,b=5π8=15π24.

Then

cos(π3)sin(5π8)=(12)(sin(23π24)sin(7π24))

=(12)(sin(π24)+sin(7π24))

where in the last line we use sin(πx)=sin(x) and sin(x)=sin(x).

As you can see, this is unwieldy compared with just putting in cos(π3)=12. The trigonometric product-sum and product-difference relations are more useful when you can't evaluate either factor in the product.

Aug 2, 2016

(12)cos(π8)

Explanation:

P=cos(π3).sin(5π8)
Trig table --> cos(π3)=12
Trig unit circle and property of complementary arcs -->
sin(5π8)=sin(π8+4π8)=sin(π8+π2)=
=cos(π8).
P can be expressed as:
P=(12)cos(π8)
NOTE. We can evaluate cos(π8) by using the trig identity:
1+cos(π4)=2cos2(π8)