How do you express #f(theta)=-cos^2(theta)-7sec^2(theta)-5csc^4theta# in terms of non-exponential trigonometric functions?
1 Answer
Explanation:
I'm not entirely sure if I have understood your question correctly - if somebody else thinks that I have misinterpreted, please let me know.
I will use the following identities for my transformations:
#sec(theta) = 1 / cos (theta)# ,#" "csc(theta) = 1 / sin(theta)#
#cos^2 (theta) = (1 + cos(2 theta))/2#
# sin^4 (theta) = (cos(4theta) - 4 cos(2 theta) + 3)/8#
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First of all, let me show you the
1a) Prove
I will use the identity
#cos(x+y) = cos(x)cos(y) - sin(x)sin(y)# Thus, it holds
#cos(2 theta) = cos (theta + theta) #
#= cos(theta)cos(theta) - sin(theta)sin(theta)#
#= cos^2(theta) - sin^2(theta)#
# = cos^2(theta) - ( 1 - cos^2(theta)) = 2 cos^2(theta) - 1 #
#<=> cos(2 theta) + 1 = 2 cos^2 (theta)#
#<=> (cos(2 theta) + 1) / 2 = cos^2 (theta)#
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1b) Prove
#sin^4(theta) = [sin^2(theta)]^2 #
#= [1 - cos^2(theta)]^2#
# = [1 - (cos(2 theta) + 1) / 2]^2#
# = 1 - 2 * (cos(2 theta) + 1)/2 + (cos^2(2 theta) + 2 cos(2 theta) + 1) / 4 #
# = 1 - cos(2 theta) - 1 + 1/4 cos^2(2 theta) + 1/2 cos(2 theta) + 1/4 #
# = 1/4 cos^2(2 theta) - 1/2 cos(2 theta) + 1/4 #
# = 1/4 (cos(4 theta) + 1) / 2 - 1/2 cos(2 theta) + 1/4 #
# = (cos(4 theta) + 1 - 4 cos(2 theta) + 2) / 8 #
# = (cos(4 theta) - 4 cos(2 theta) + 3) / 8 #
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2) Apply the identities
So, now I will apply the identities from above:
#f(theta) = - cos^2(theta) - 7 sec^2(theta) - 5 csc^4(theta)#
# = - cos^2(theta) - 7 / cos^2 (theta) - 5 / sin^4 (theta) #
# = - (cos(2 theta) + 1) / 2 - (7*2) / (cos (2 theta) + 1) - (5 * 8) / (cos(4 theta) - 4 cos(2 theta) + 3) #
# = - (cos(2 theta) + 1) / 2 - 14 / (cos (2 theta) + 1) - 40 / (cos(4 theta) - 4 cos(2 theta) + 3) #
So... I hope that this helped at all. The final expression is non-exponential and contains only the
However, I don't know if you mind having trigonometric functions in the denominator or having more than one fraction. If yes, please let me know and I'll see if I can think of a different solution.