I'm relearning highschool mathematics, and teaching myself higher level mathematics, and I spent a few minutes on this question but got stuck at the point of determining cos(6b) = ? Please look at the question below, and correct my mistakes and determine the final answer. I'm curious. I have no homework to turn in, but I'd love to know how this works.
Thank you!
http://answers.yahoo.com/question/index;_ylt=AoQp4icTYHdSMTAd_9Lei3bty6IX;_ylv=3?qid=20071104012803AAnkttq&show=7#profile-info-uDLXZyxAaa
2007-11-04 18:48:46 · 2 answers · asked by rachelesse 3 in Science & Mathematics ➔ Mathematics
The question is:
angle p which lies between pi and 3pi/2 and which satisfies sin(p)=6(b).(b constant).what is the value of cos^-1(6b) and cos(6b)(express it in terms of p and/or b)
_____________
sin(p) = 6b
arcsin(6b) = p
cos(p) = √[1 - sin²p] = √[1 - (6b)²] = √(1 - 36b²)
[arcsin(6b) + arccos(6b)]/2 = [π + 3π/2]/2 = 5π/4
arcsin(6b) + arccos(6b) = 5π/2
arccos(6b) = 5π/2 - arcsin(6b)
2007-11-04 19:12:07 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
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2007-11-05 03:03:00 · answer #2 · answered by arshad 2 · 0⤊ 4⤋