Can anyone help me condense the expression cos(2alpha-beta) so that it only has one of the given terms?

Question

What I am looking to achieve here is to use more than one trigonometric identities to write cos(2alpha-beta) as an expression that either only concerns alpha or beta, and not both. I have been stuck and am not sure how to progress besides potentially using a double-angle formula. I was attempting to only use alpha, but was not quite sure how to get it to where I could get rid of beta.

Thank you in advance!

Answer 1

If alpha and beta are arbitrary angles, you cannot eliminate information about one of them.

Answer 2

Given the additional information:

tan(alpha) = -20/21 where alpha is in Quadrant II

csc(beta) = -61/11 where beta is in Quadrant III

Using the identity 1+(tan (alpha))²=(sec (alpha))², we have |sec (alpha)| = 29/21

In Quadrant II, the cosine is negative and the sine is positive

Since cos (alpha) = 1/sec (alpha), we have cos (alpha) = -21/29 and sin (alpha) = 20/29

sin (beta) =1/csc (beta) = -11/61

In Quadrant III, the cosine is negative

so cos (beta) = -√(1-(sin (beta))² = -60/61

What is the value of cos (2 alpha - beta)?

cos (2 alpha - beta) = cos (2 alpha) cos (beta) + sin (2 alpha) sin (beta)

cos (2 alpha) = 2 (cos (alpha))^2 - 1 = 41/841

sin (2 alpha) = 2 sin (alpha) cos (alpha) = -840/841

and from there on, all you need is a calculator