Sin(a+b)sin(a-b) = cos^2b - cos^2a

  • Thread starter zeion
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In summary, the homework statement is trying to solve for cos2b using the sum and difference identities. After expanding the left side, the proof becomes uncomplicated.
  • #1
zeion
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1

Homework Statement


Prove that: sin(a+b)sin(a-b) = cos^2b - cos^2a


Homework Equations





The Attempt at a Solution



I'm stuck.
I've tried expanding the left side, making the right side to a difference of squares, subbing the right side with sin's, dividing, factoring and all that stuff nothing gets me anywhere..

Any hints?
 
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  • #2
After using the sum and difference identities, expand the left side. You should get a difference of two squares. One of the terms will contain a sin2 b. Replace this with an expression using a well-known trig identity. Eventually, you will get the right side. If you're stuck, show us what you got.
 
  • #3
After you expand the left side you should have
sin2(a)cos2(b) - cos2(a)sin2(b)

Replace both sine expressions with their cosine counterparts.
 
  • #4
After trying this myself, the proof seems uncomplicated.

The lefthand side immediately becomes (sin(a)cos(b)+cos(a)sin(b))(sin(a)cos(b)-cos(a)sin(b)),
= (sin(a)cos(b))2-(cos(a)sin(b))2
=sin2a cos2b - cos2a sin2b

Next we use our most basic circular identity and set up and substitute for sin2a=1-cos2a, and similar for sin2b=1-cos2b

Giving lefthandside = (1-cos2a)cos2b-cos2a(1-cos2b)

...from which you can continue the algebra and find the righthand side expression which you expect.
 
  • #5
symbolipoint said:
After trying this myself, the proof seems uncomplicated.

For you, maybe. ~_~

Thanks guys.
 
  • #6
zeion said:
For you, maybe. ~_~

Thanks guys.

Don't listen to him, he's being all in your face hardcore with his maths guns :-p
 
  • #7
About #5 and #6:
At least I showed how to start and how to progress. I did not finish the algebra steps on the forum; I left those for zeion.
 
  • #8
could someone show how to do the algebra? I've gotten to

(1-cos2(a))(cos2(b)) - (cos2(a))(1-cos2(b)) = (1-cos2(a)) - (1-cos2(b))

but for some reason the algebra just isn't making sense to me?
 
  • #9
claireashley3 said:
could someone show how to do the algebra? I've gotten to

(1-cos2(a))(cos2(b)) - (cos2(a))(1-cos2(b)) = (1-cos2(a)) - (1-cos2(b))

but for some reason the algebra just isn't making sense to me?
At first glance, I'm not sure how you got the right hand side. Just distribute the left hand side.
[tex](1 - \cos^2 a)(\cos^2 b) - (\cos^2 a)(1 - \cos^2 b)[/tex]
[tex]= \cos^2 b - \cos^2 a \cos^2 b - \cos^2 a + \cos^2 a \cos^2 b[/tex]
Do you see it now?
 
  • #10
(1-cos2(a))-(1-cos2(b))

= 1-cos2(a)-1+cos2(b)

Right?
 
  • #11
ah yeah thanks sorry
looking at it now it makes sense, i messed up the right side
thank you!
 

FAQ: Sin(a+b)sin(a-b) = cos^2b - cos^2a

What is the formula for "Sin(a+b)sin(a-b) = cos^2b - cos^2a"?

The formula is a trigonometric identity that expresses the product of the sums and differences of two angles, a and b, in terms of the cosine of those angles.

How is this formula derived?

This formula can be derived using the sum and difference identities for sine and cosine, as well as the Pythagorean identity for cosine.

What is the significance of this formula?

This formula is useful for simplifying and solving trigonometric equations, as well as for proving other identities.

Can this formula be used to find values of sine and cosine?

No, this formula does not directly give the values of sine and cosine, but it can be used to simplify expressions involving sine and cosine.

Are there any other equivalent forms of this formula?

Yes, this formula can also be written as cos(a+b)cos(a-b) = sin^2b - sin^2a, or in terms of tangent as tan(a+b)tan(a-b) = 1 - tan^2a.

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