__maths__>__Advanced Trigonometry__>__Trigonometric Identities for compound angles__### Compound Angles: cos(A+B), sin(A-B), cos(A-B)

In this page, a simple proof, based on the earlier result of `sin(A+B)`, is explained for expressing `cos(A+B)`, `sin(A-B)`, `cos(A-B)` in terms of `sinA`, `sinB`, `cosA` , and `cosB`.

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It was geometrically proven that `cos(A+B)=cosA cosB - sinA sinB`. For the same result, we can work out a proof using algebra of trigonometric functions.

`cos(A+B)`

`quad quad = sin(90-A-B)`

`quad quad = sin[(90-A) + (-B)]`

`quad quad = sin(90-A)cos(-B)+cos(90-A)sin(-B)`

`quad quad = cosA cosB - sinA sinB`

To calculate trigonometric values for compound angle, we will switch to using the proofs with algebra of trigonometric functions, as it is simpler. But, equivalently a geometrical proof can be worked out.

Proof for sin(A-B) and cos(A-B) using previous results and algebra of trigonometric functions.

`sin(A-B)`

`quad quad = sin(A+(-B))`

`quad quad = sinA cos(-B)+ cosA sin(-B)`

`quad quad = sinA cosB - cosA sinB`

`cos(A-B)`

`quad quad = cos(A+(-B))`

`quad quad = cosA cos(-B) - sinA sin(-B)`

`quad quad = cosA cosB + sinA sinB`

Geometrical proof for `sin(A-B)` and `cos(A-B)` is outlined below. equate square of chord lengths to derive the result given.

`bar(RT)^2 = bar(PQ1)^2`

`=> sin(A-B) = sinA cosB - cosA sinB `

`bar(RS)^2 = bar(PQ)^2`

`=> cos(A-B) = cosA cosB + sinA sinB`

`sin(A-B)=sinA cosB - cosA sinB`

`cos(A-B)=cosA cosB + sinA sinB`

*Solved Exercise Problem: *

Compute `sin75`.

- cannot compute as `75^@` is not a standard angle
- consider `75^@` as sum of standard angles `45^@` and `30^@`
- consider `75^@` as sum of standard angles `45^@` and `30^@`

The answer is 'consider `75^@` as sum of standard angles `45^@` and `30^@`'.

`sin75`

`quad = sin(45+30)`

`quad = sin45cos30 + cos45sin30`

`quad = (1+ sqrt(3))/(2sqrt(2))`

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