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

2 Answers
Oscar L.
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.

Nghi N.
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)

Related questions
See all questions in Products, Sums, Linear Combinations, and Applications
Impact of this question
2353 views around the world
You can reuse this answer
Creative Commons License