# 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.