How do you simplify #f(theta)=-tan4theta+sin2theta+cos4theta# to trigonometric functions of a unit #theta#?

1 Answer
Jun 10, 2018

#-(4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)#
#qquadqquad -2sin theta cos theta+1-8sin^2theta cos^2theta#

Explanation:

We start with the double angle identities :

#sin(2 theta) = 2 sin theta cos theta#

and

#cos(2 theta) = cos^2 theta - sin^2 theta = 2cos^2 theta-1 = 1-2sin^2 theta#

Now

#cos (4 theta) = cos(2times 2theta)#
#qquadqquadqquad = 1-2 sin^2(2theta)#
#qquadqquadqquad = 1-2(2 sin theta cos theta)^2#
#qquadqquadqquad = 1-8sin^2theta cos^2theta#

and

#sin(4 theta) = sin(2 times 2theta)#
#qquadqquadqquad = 2 sin(2theta)cos(2theta)#
#qquadqquadqquad = 2(2 sin theta cos theta)(1-2sin^2theta)#
#qquadqquadqquad = 4 sin theta cos theta (1-2 sin^2 theta)#

leading to

#tan(4theta) = sin(4 theta)/cos(4 theta)#
#qquadqquadqquad = (4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)#

Thus

#f(theta) = -tan(4 theta)+sin(2 theta)+cos(4 theta)#
#qquad\ quad=-(4 sin theta cos theta (1-2 sin^2 theta))/(1-8sin^2theta cos^2theta)#
#qquadqquad -2sin theta cos theta+1-8sin^2theta cos^2theta#