Find the solution of cos(α−β)

Question

Given cos(α)=−√17/9 and π/2<α<π and cos(β)=−√24/7 and β is in quadrant III. Use sum and difference formulas to find the following:

Note: You are not allowed to use decimals in your answer.

cos(α−β)=

Answer 1

A is in quadrant II

B is in quadrant III

cosA and cosB are negative

sinA >0, sinB<0

cosA =-sqr17/9

use the Pythagorean Theorem

sinA = sqr(9^2-17)/9 = sqr64/9 = 8/9

cosB =-sqr24/7,

sinB = -sqr(7^2-24)/7 = -sqr25/7 =-5/7

difference formula for consines is:

cos(A-B) = cosAcosB + sinAsinB

plug in the values for cosines and sines

= [(-sqr17)/9][(-sqr24)/7] + [[8/9][-5/7]

= sqr[(24)(17)]/9(7) - 40/63

= [sqr408 -40]/63

= [2(sqr102) -40]/63

as simple as it gets for an exact answer

check the answer

A = about 117.3

B = about 225.6

A-B = -108.3

cos(A-B) =cos(-108.3) = about -0.3

(2sqr102 -40)/63 = about -0.3