Trig: product as a sum?

Question

Ok, I am really lost on this one and I have to solve this (as well as some others) by today, please help.


It says to "Express the product as a sum containing only sines or cosines. Your answer must be simplified completely."

sin(Θ/2)cos(7Θ/2)

I'm just lost from here. Can someone help me to solve this or point me towards a formula or something I should use? I really appreciate it, thanks!

Answer 1

I'm going to use x instead of theta.
sin (x/2) cos (7x/2)

There is an identity:
sin a cos b = (1/2)[sin (a + b) + sin (a - b)]

in our case...
a = x/2
b = 7x/2

so...
sin (x/2) cos (7x/2)
= (1/2)[sin (x/2 + 7x/2) + sin (x/2 - 7x/2)]
= (1/2)[sin (8x/2) + sin (-6x/2)]
= (1/2)[sin 4x + sin (-3x)]

Note:
sin (-a) = - sin a

(1/2)[sin 4x + sin (-3x)]
= (1/2)[sin 4x - sin 3x]

There are more identities listed in the website below, if you need to simplify it further.

Answer 2

Using the Product-to-sum equation for sin*cos you end up with Sin(4*Theta) + Sin (-3*Theta) all over 2.

and i think that can't be reduced after that.

Answer 3

It appears your professor on some absent mind or on busy with some other assignments slightly erred in providing the correct answer. As well you also had not simplified it fully. Actually the correct simplified answer is: {-2 Cos(7x/2)*Sin(3x/2)} For suiting with the convenience of the formula, better you convert the problem as: -{Sin(5x) - Sin(2x)}; then you will get without any difficulty the answer in simplified form.

Answer 4

sinA-sinB= 2sin(1/2(A-B))cos(1/2(A+B))
so A-B= a A+B= 7a A= 4a B=3a

1/2(sin 4Θ) -1/2(sin 3Θ)= sin(Θ/2)cos(7Θ/2)

Answer 5

There is an identity you need:

sin(x) - sin(y) = 2sin((x-y)/2)*cos((x+y)/2)

Now if x = 4*theta and y = 3*theta then

sin(4 theta) - sin(3 theta) = 2 sin(theta/2)*cos(7theta/2)

So
sin(theta/2)cos(7theta/2) = 1/2( sin(4theta)-sin(3theta))

Answer 6

Use the formula
sinAcosB = 1/2 ( sin(A-B) + sin(A+B))

sin(Θ/2)cos(7Θ/2) =1/2 (sin(-3Θ) + sin(4Θ))
=1/2(sin 4Θ -sin 3Θ)