Question

How do you use the product to sum formulas to write `4cos(pi/3)sin((5pi)/6)` as a sum or difference?

Answer 1

`1.`

Explanation:

`4cos(pi/3)sin(5pi/6)=4cos(pi/3)sin(pi-pi/6),`

`=4cos(pi/3)sin(pi/6),`

`=2{2cos(pi/3)sin(pi/6)},`

`=2{sin(pi/3+pi/6)-sin(pi/3-pi/6)},`

`=2{sin(pi/2)-sin(pi/6)},`

`=2(1-1/2),`

`=1.`

Answer 2

`2[cos ((2pi)/3) - cos pi]`

Explanation:

`y = 4cos (pi/3)sin ((5pi)/6)`
Because,
`cos (pi/3) = sin (pi/2 - pi/3) = sin (pi/6)`
Therefor,
`y = 4sin (pi/6).sin ((5pi)/6)`
Apply the trig identity:
`sin a.sin b = (1/2)[cos (a - b) - cos (a + b)]`
In this case:
`y = 2[cos ((2pi)/3) - cos pi]`
That is the answer. However, we can go further, knowing
`cos ((2pi)/3) = - 1/2` and `cos pi = - 1`.
`y = 2(- 1/2 + 1) = 2/2 = 1`