How do you solve #sin 4beta = cos 5beta#?

1 Answer
Nghi N.
Aug 18, 2016

#pi/18 + 2kpi#
#(3pi)/2 + 2kpi#

Explanation:

sin 4x = cos 5x (1)
Use trig identity:
#sin a = cos (pi/2 - a)#
#sin 4x = cos (pi/2 - 4x)#
From (1):
#cos 5x = cos (pi/2 - 4x)#
#5x = +- (pi/2 - 4x)#
a. #5x = pi/2 - 4x #--> #9x = pi/2# --> #x = pi/18#
b. #5x = - pi/2 + 4x# --> #x = - pi/2#, or #x = (3pi)/2#
General answer
#x = pi/18 + 2kpi#
#x = (3pi)/2 + 2kpi#
Check
#x = pi/18 = 10^@# --> sin 4x = sin 40 = 0.642
cos 5x = cos 50 = 0.642. OK

Related questions
See all questions in Relating Trigonometric Functions
Impact of this question
2488 views around the world
You can reuse this answer
Creative Commons License