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

Jan 25, 2016

$f \left(\theta\right) = - \frac{\cos \left(2 \theta\right) + 1}{2} - \frac{14}{\cos \left(2 \theta\right) + 1} - \frac{40}{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}$

#### 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:

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$, $\text{ } \csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$

${\sin}^{4} \left(\theta\right) = \frac{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}{8}$

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First of all, let me show you the ${\cos}^{2} \left(\theta\right)$ and ${\sin}^{4} \left(\theta\right)$ identities:

1a) Prove ${\cos}^{2} \left(\theta\right) = \frac{\cos \left(2 \theta\right) + 1}{2}$

I will use the identity

$\cos \left(x + y\right) = \cos \left(x\right) \cos \left(y\right) - \sin \left(x\right) \sin \left(y\right)$

Thus, it holds

$\cos \left(2 \theta\right) = \cos \left(\theta + \theta\right)$

$= \cos \left(\theta\right) \cos \left(\theta\right) - \sin \left(\theta\right) \sin \left(\theta\right)$

$= {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

$= {\cos}^{2} \left(\theta\right) - \left(1 - {\cos}^{2} \left(\theta\right)\right) = 2 {\cos}^{2} \left(\theta\right) - 1$

$\iff \cos \left(2 \theta\right) + 1 = 2 {\cos}^{2} \left(\theta\right)$

$\iff \frac{\cos \left(2 \theta\right) + 1}{2} = {\cos}^{2} \left(\theta\right)$

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1b) Prove ${\sin}^{4} \left(\theta\right) = \frac{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}{8}$

${\sin}^{4} \left(\theta\right) = {\left[{\sin}^{2} \left(\theta\right)\right]}^{2}$

$= {\left[1 - {\cos}^{2} \left(\theta\right)\right]}^{2}$

$= {\left[1 - \frac{\cos \left(2 \theta\right) + 1}{2}\right]}^{2}$

$= 1 - 2 \cdot \frac{\cos \left(2 \theta\right) + 1}{2} + \frac{{\cos}^{2} \left(2 \theta\right) + 2 \cos \left(2 \theta\right) + 1}{4}$

$= 1 - \cos \left(2 \theta\right) - 1 + \frac{1}{4} {\cos}^{2} \left(2 \theta\right) + \frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{4}$

$= \frac{1}{4} {\cos}^{2} \left(2 \theta\right) - \frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{4}$

$= \frac{1}{4} \frac{\cos \left(4 \theta\right) + 1}{2} - \frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{4}$

$= \frac{\cos \left(4 \theta\right) + 1 - 4 \cos \left(2 \theta\right) + 2}{8}$

$= \frac{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}{8}$

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2) Apply the identities

So, now I will apply the identities from above:

$f \left(\theta\right) = - {\cos}^{2} \left(\theta\right) - 7 {\sec}^{2} \left(\theta\right) - 5 {\csc}^{4} \left(\theta\right)$

$= - {\cos}^{2} \left(\theta\right) - \frac{7}{\cos} ^ 2 \left(\theta\right) - \frac{5}{\sin} ^ 4 \left(\theta\right)$

$= - \frac{\cos \left(2 \theta\right) + 1}{2} - \frac{7 \cdot 2}{\cos \left(2 \theta\right) + 1} - \frac{5 \cdot 8}{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}$

$= - \frac{\cos \left(2 \theta\right) + 1}{2} - \frac{14}{\cos \left(2 \theta\right) + 1} - \frac{40}{\cos \left(4 \theta\right) - 4 \cos \left(2 \theta\right) + 3}$

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.