How do you express #f(theta)=sin^2(theta)-3cot^2(theta)+csc^4theta# in terms of non-exponential trigonometric functions?

1 Answer
Lotusbluete
Feb 2, 2016

#f(theta) = (1 - cos(2theta)) / 2 + ( 5 + 3 cos(4theta) )/( 3 - 4cos(2theta) + cos(4theta) )#

Explanation:

Let me try to do that.

I hope that I have understood your question correctly - if anybody thinks that I have misinterpreted something, please let me know!

I will use the following identities:

#csc (theta) = 1 / sin(theta)#, #color(white)(xxxx) cot (theta) = cos(theta) / sin(theta)#

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

#cos^2(theta) = (1 + cos(2theta)) / 2#

=======================

First of all, let me show the #sin^2(theta)# and #cos^2(theta)# identities:

1a) Prove #" "sin^2(theta) = (1 - cos(2theta)) / 2#

I will use the identities

#[1] color(white)(xxx) cos(x + y) = cos(x)cos(y) - sin(x) sin(y)#

#[2] color(white)(xxx) sin^2 (x) + cos^2 (x) = 1#

Thus, it holds

# cos(2 theta) = cos(theta + theta)#

#color(white)(xxxxx) stackrel([1])(=) cos(theta)cos(theta) - sin(theta)sin(theta)#

#color(white)(xxxxx) = cos^2(theta) - sin^2(theta)#

#color(white)(xxxxx) stackrel([2])(=) (1 - sin^2(theta)) - sin^2(theta)#

#color(white)(xxxxx) = 1 - 2 sin^2 theta#

#color(white)(xx)#
#<=> cos(2 theta) - 1 = - 2 sin^2 (theta)#

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

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

=======================

1b) Prove #" "cos^2(theta) = (1 + cos(2theta)) / 2#

The proof is very similar to the one above:

# cos(2 theta) stackrel([1])(=) cos(theta)cos(theta) - sin(theta)sin(theta)#

#color(white)(xxxxx) = cos^2(theta) - sin^2(theta)#

#color(white)(xxxxx) stackrel([2])(=) cos^2(theta) - (1 - cos^2(theta))#

#color(white)(xxxxx) = 2 cos^2(theta) - 1#

#<=> cos(2 theta) + 1 = 2 cos^2(theta)#

#<=> (cos(2 theta) + 1) / 2 = cos^2(theta) #

=======================

2) Apply the identities

So, now I will apply the identities and simplify:

#sin^2(theta) - 3 cot^2(theta) + csc^4(theta)#

# = sin^2(theta) - 3 [cos(theta)/sin(theta)]^2 + 1 / sin^4(theta)#

# = sin^2(theta) - (3cos^2(theta))/sin^2(theta) + 1 / [sin^2(theta)]^2#

... combine the two latter fractions into one by determining the least common denominator...

# = sin^2(theta) - (3cos^2(theta) * sin^2(theta))/(sin^2(theta) * sin^2(theta)) + 1 / [sin^2(theta)]^2#

# = sin^2(theta) + (- 3cos^2(theta)sin^2(theta) + 1)/[sin^2(theta)]^2 #

... apply the #sin^2(theta)# and #cos^2(theta)# identities...

# = (1 - cos(2theta)) / 2 + (-3* (1 + cos(2theta)) / 2 * (1 - cos(2theta)) / 2 + 1)/([(1 - cos(2theta)) / 2]^2)#

# = (1 - cos(2theta)) / 2 + (-3* ((1 + cos(2theta))* (1 - cos(2theta))) / 4 + 4/4)/((1 - cos(2theta))^2 / 4)#

... get rid of the double fractions by factoring #1/4# both in the numerator and the denominator and then canceling...

# = (1 - cos(2theta)) / 2 + ( cancel(1/4) * (-3(1 + cos(2theta)) * (1 - cos(2theta)) + 4) )/(cancel(1/4) * (1 - cos(2theta))^2 )#

# = (1 - cos(2theta)) / 2 + ( -3(1 + cos(2theta)) * (1 - cos(2theta)) + 4 )/(1 - cos(2theta))^2#

... use the formulas #(a+b)(a-b) = a^2 - b^2# and #(a-b)^2 = a^2 - 2ab + b^2#...

# = (1 - cos(2theta)) / 2 + ( -3(1^2 - cos^2(2theta)) + 4 )/(1^2 - 2cos(2theta) + cos^2(2 theta))#

# = (1 - cos(2theta)) / 2 + ( 1 + 3 cos^2(2 theta) )/(1 - 2cos(2theta) + cos^2(2 theta))#

... apply the #cos^2(theta)# identity again...

# = (1 - cos(2theta)) / 2 + ( 1 + 3 * (1 + cos(4theta)) / 2 )/(1 - 2cos(2theta) + (1 + cos(4theta)) / 2)#

... again, get rid of the double fractions by factoring #1/2# both in the numerator and the denominator and then canceling...

# = (1 - cos(2theta)) / 2 + ( cancel(1/2) (2 + 3 (1 + cos(4theta)) ) )/( cancel(1/2) (2 - 4cos(2theta) + (1 + cos(4theta)) ))#

# = (1 - cos(2theta)) / 2 + ( 2 + 3 (1 + cos(4theta)) )/( 2 - 4cos(2theta) + (1 + cos(4theta)) )#

# = (1 - cos(2theta)) / 2 + ( 5 + 3 cos(4theta) )/( 3 - 4cos(2theta) + cos(4theta) )#

This is now a non-exponential expression that uses only the #cos# function.

Hope that this helped! :-)

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