Question

How do you express `f(theta)=-cos^2(theta)-7sec^2(theta)-5csc^4theta` in terms of non-exponential trigonometric functions?

Answer 1

`f(theta) = - (cos(2 theta) + 1) / 2 - 14 / (cos (2 theta) + 1) - 40 / (cos(4 theta) - 4 cos(2 theta) + 3)`

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 `cos^2(theta)` and `sin^4 (theta)` identities:

1a) Prove `cos^2 (theta) = (cos(2 theta) + 1)/2`

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) = (cos(4theta) - 4 cos(2 theta) + 3)/8`

`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 `cos` function.

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.