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

Feb 2, 2016

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

#### 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:

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$, $\textcolor{w h i t e}{\times \times} \cot \left(\theta\right) = \cos \frac{\theta}{\sin} \left(\theta\right)$

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

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

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

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

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

I will use the identities

$\left[1\right] \textcolor{w h i t e}{\times x} \cos \left(x + y\right) = \cos \left(x\right) \cos \left(y\right) - \sin \left(x\right) \sin \left(y\right)$

$\left[2\right] \textcolor{w h i t e}{\times x} {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Thus, it holds

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

$\textcolor{w h i t e}{\times \times x} \stackrel{\left[1\right]}{=} \cos \left(\theta\right) \cos \left(\theta\right) - \sin \left(\theta\right) \sin \left(\theta\right)$

$\textcolor{w h i t e}{\times \times x} = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

$\textcolor{w h i t e}{\times \times x} \stackrel{\left[2\right]}{=} \left(1 - {\sin}^{2} \left(\theta\right)\right) - {\sin}^{2} \left(\theta\right)$

$\textcolor{w h i t e}{\times \times x} = 1 - 2 {\sin}^{2} \theta$

$\textcolor{w h i t e}{\times}$
$\iff \cos \left(2 \theta\right) - 1 = - 2 {\sin}^{2} \left(\theta\right)$

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

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

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

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

The proof is very similar to the one above:

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

$\textcolor{w h i t e}{\times \times x} = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

$\textcolor{w h i t e}{\times \times x} \stackrel{\left[2\right]}{=} {\cos}^{2} \left(\theta\right) - \left(1 - {\cos}^{2} \left(\theta\right)\right)$

$\textcolor{w h i t e}{\times \times x} = 2 {\cos}^{2} \left(\theta\right) - 1$

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

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

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

2) Apply the identities

So, now I will apply the identities and simplify:

${\sin}^{2} \left(\theta\right) - 3 {\cot}^{2} \left(\theta\right) + {\csc}^{4} \left(\theta\right)$

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

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

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

$= {\sin}^{2} \left(\theta\right) - \frac{3 {\cos}^{2} \left(\theta\right) \cdot {\sin}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right) \cdot {\sin}^{2} \left(\theta\right)} + \frac{1}{{\sin}^{2} \left(\theta\right)} ^ 2$

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

... apply the ${\sin}^{2} \left(\theta\right)$ and ${\cos}^{2} \left(\theta\right)$ identities...

$= \frac{1 - \cos \left(2 \theta\right)}{2} + \frac{- 3 \cdot \frac{1 + \cos \left(2 \theta\right)}{2} \cdot \frac{1 - \cos \left(2 \theta\right)}{2} + 1}{{\left[\frac{1 - \cos \left(2 \theta\right)}{2}\right]}^{2}}$

$= \frac{1 - \cos \left(2 \theta\right)}{2} + \frac{- 3 \cdot \frac{\left(1 + \cos \left(2 \theta\right)\right) \cdot \left(1 - \cos \left(2 \theta\right)\right)}{4} + \frac{4}{4}}{{\left(1 - \cos \left(2 \theta\right)\right)}^{2} / 4}$

... get rid of the double fractions by factoring $\frac{1}{4}$ both in the numerator and the denominator and then canceling...

$= \frac{1 - \cos \left(2 \theta\right)}{2} + \frac{\cancel{\frac{1}{4}} \cdot \left(- 3 \left(1 + \cos \left(2 \theta\right)\right) \cdot \left(1 - \cos \left(2 \theta\right)\right) + 4\right)}{\cancel{\frac{1}{4}} \cdot {\left(1 - \cos \left(2 \theta\right)\right)}^{2}}$

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

... use the formulas $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ and ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$...

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

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

... apply the ${\cos}^{2} \left(\theta\right)$ identity again...

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

... again, get rid of the double fractions by factoring $\frac{1}{2}$ both in the numerator and the denominator and then canceling...

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

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

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

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

Hope that this helped! :-)