cosalpha = 1/5_____ sinalpha = 2sqrt(6)/5 _____ tanalpha = 2sqrt(6)
cosbeta = 2/5______ sinbeta = sqrt(21)/5 ______ tanbeta = sqrt(21)/2
cosgamma = 3/5___ singamma = 4/5 _________ tangamma = 4/3
cosdelta = 4/5 _____sindelta = 3/5 ___________ tandelta = 3/4
cos(alpha + beta) = (2 - 6sqrt(14))/25
sin(alpha + beta) = (sqrt(21) + 4sqrt(6))/25
tan(alpha + beta) = 50(sqrt(21)+sqrt(6))/-500 = -(sqrt(21)+sqrt(6))/10
Corrected from original. Thanks to Laura Hidrobo.
cos(beta - gamma) = (6 + 4sqrt(21))/25
sin(beta - gamma) = (3sqrt(21) - 8)/25
tan(beta - gamma) = 1 - sqrt(21)/6
Corrected from original. Thanks to Bi Ting Zhen and Laura Hidrobo.
cos(gamma + delta) = 0
sin(gamma + delta) = 1
tan(gamma + delta) =undefined
The last one is this way because the angles are complementary.
Hi, can you show me how you got the tan for the first and 2nd on the lower half of the quiz please. I am not getting your answer but instead the 25 in the demoninators cancel, and I am getting -500 in the denominator. In the numerator sqrt(21)+sqrt(6))/-10. The answer must be negative as well. Im confused with this.
ReplyDeleteThanks, Laura. You were right and I was wrong. I've fixed the post and given you two points extra credit on homework 11.
ReplyDeleteHi, I have a question about the 2nd tangent problem. I am getting (negative sqrt 21)/6 +1 your getting a positive. Where am I wrong?
ReplyDeleteHi, Laura. I went too fast again. The simplest way to write it is 1 - sqrt(21)/6.
ReplyDeleteOne more extra credit point for you.
Thanks :) I was reviewing it and reviewing it LOL
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDelete