If #sinx-cosx=1/3#, find #cos(4x)#?

2 Answers
Dec 6, 2017

cos 4x = - 0.58
#x = 58^@63#; and #x = 211^@37#

Explanation:

Use trig identity:
#sin x - cos x = sqrt2sin (x - pi/4)#
In this case
#sqrt2sin (x - pi/4) = 1/3#
#sin (x - pi/4) = 1/(3sqrt2) = sqrt2/6#
Calculator and unit circle give -->

a. x - 45 = 13.63 --> #x = 45 + 13.63 = 58^@63#
4x = 234^@52 --> #cos 4x = cos 234^@52 = - 0.58#

b. x - 45 = 180 - 13.63 = 166.37
#x = 166.37 + 45 = 211^@37#
#4x = 845.48 = 845.48 - 720 = 125^@48#
#cos 4x = cos 125^@48 = - 0.58#
Check by calculator.
x = 58.63 --> sin x = 0.85 --> cos x = 0.52.
sin x - cos x = 0.33 = 1/3. Proved
x = 211.37 --> sin x = - 0.52 --> cos x = - 0.85
sin x - cos x = - 0.52 - (-0.85) = 0.33 = 1/3. Proved

#cos(4x) = - 47/81#

Explanation:

We have #sinx-cosx=1/3#

#sin^2x + cos^2x - 2sinxcosx = 1/9#

#1 - 1/9 = sin(2x)#

#8/9 = sin(2x)#

We know that

#sin^2(2x) + cos^2(2x) = 1#

#(8/9)^2 + cos^2(2x) = 1#

#cos^2(2x) =1 - 64/81#

#cos^2(2x) = 17/81#

We can rewrite

#cos(4x)# as #2cos^2(2x) - 1#

#cos(4x) = 2xx17/81 - 1#

#cos(4x) = - 47/81#