How do you express #f(theta)=sin^2(theta)-2sec^2(theta)-cot^2theta# in terms of non-exponential trigonometric functions?

1 Answer
Apr 8, 2018

#f(theta) = 1/4 (-42+15cos(2theta)-6cos(4theta)+cos(6theta))/(1-cos(4theta))#

Explanation:

#f(theta)=sin^2(theta)-2sec^2(theta)-cot^2theta#

#qquad = sin^2(theta)-2/cos^2(theta)-cos^2(theta)/sin^2(theta)#

#qquad = (sin^4(theta)cos^2(theta)-2sin^2(theta)-cos^4(theta))/(cos^2(theta)sin^2(theta)#

Now we us the double angle formulas

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

and

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

to get

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

#cos^4(theta) = (1/2(1+cos(2theta)))^2#
#qquad = 1/4(1+2cos(2theta)+cos^2(2theta))#
#qquad = 1/4+1/2cos(2theta)+1/4(1/2(1+cos(4theta)))#
#qquad = 3/8+1/2 cos(2theta)+1/8 cos(4theta)#,

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

and

#sin^4(theta) cos^2(theta)= sin^2(theta)times (sin^2(theta)cos^2(theta))#
#qquad = 1/2(1-cos(2theta))times 1/8(1-cos(4theta))#
#qquad 1/16 (1-cos(2theta)-cos(4theta)+cos(2theta)cos(4theta))#

Now, we use the formula

#2cosAcosB = cos(A+B)+cos(A-B)#

to rewrite the last term to get

#sin^4(theta) cos^2(theta)= 1/16 (1-cos(2theta)-cos(4theta)+1/2cos(6theta)+1/2cos(2theta))#
#qquad = 1/32 (2-cos(2theta)-2cos(4theta)+cos(6theta))#

Combining all these we get

#sin^4(theta)cos^2(theta)-2sin^2(theta)-cos^4(theta)#
#qquad = 1/32 (2-cos(2theta)-2cos(4theta)+cos(6theta))#
#qquad qquad -1+cos(2theta)#
#qquad qquad -(3/8+1/2 cos(2theta)+1/8 cos(4theta))#
#qquad = 1/32 (-42+15cos(2theta)-6cos(4theta)+cos(6theta))#

Thus, finally

#f(theta) = 1/4 (-42+15cos(2theta)-6cos(4theta)+cos(6theta))/(1-cos(4theta))#