Hello, can anyone tell me how to convert #cosα-cos3 α# to #2cos αcosbeta#? Thank you :)

Question

I did this,but the answer in my book is different.

817 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-20T12:42:21

See explanation.

#cos alpha - cos 3alpha = 2 sin (1/2(alpha + 3alpha)) sin ( 1/2( 3alpha - alpha ))#

#= 2 sin 2alpha sin alpha#

#= 4 sin alpha cos alpha sin alpha#

#= 2cos alpha (2 sin^2alpha)#

Assuming that #2 sin^2alpha <= 1#

# rArr abs( sin alpha) <= 1/sqrt2 rArr alpha in [2kpi - pi/4, 2kpi + pi/4],#

#k = 0, +-1, +-2, +-3, ...#. Then, upon setting

#beta = 2kpi+- cos^(-1)(2 sin^2alpha)#, so that

#cos beta = cos (2kpi+- cos^(-1)(2 sin^2alpha))#

#= cos ( cos^(-1)(2 sin^2alpha)) = 2 sin^2alpha#, Now,

#cos alpha - cos 3alpha = 2 cos alpha cos beta#..

For example, if #alpha = pi/6, cos beta = .1/2 = cos (pi/3)# and

#beta = 2kpi +- pi/3, k = 0, +-1, +-2, +-3, ...#

#=+-pi/3, +-5/3pi, +- 7/3pi, ...#