# How do you express f(theta)=sin^2(theta)+6cot^2(theta)-8cos^4theta in terms of non-exponential trigonometric functions?

Jul 10, 2016

We know that, $\cos 2 x = 2 {\cos}^{2} x - 1 = 1 - 2 {\sin}^{2} x$

From these, we get, ${\sin}^{2} x = \frac{1 - \cos 2 x}{2} , \ldots \ldots \ldots . . \left(1\right)$ and, ${\cos}^{2} x = \frac{1 + \cos 2 x}{2.} \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

Now, ${\cos}^{4} x = {\left({\cos}^{2} x\right)}^{2} = {\left\{\frac{1 + \cos 2 x}{2}\right\}}^{2} = \frac{1}{4} \left\{1 + 2 \cos 2 x + {\cos}^{2} \left(2 x\right)\right\}$

Here, to convert ${\cos}^{2} \left(2 x\right)$ into multiple angle, we will reuse $\left(2\right) ,$ and, put $2 x$ in place of $x ,$ to get,

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

Therefore, ${\cos}^{4} x = \frac{1}{4} \left\{1 + 2 \cos 2 x + \frac{1 + \cos 4 x}{2}\right\} = \frac{1}{8} \left(3 + 4 \cos 2 x + \cos 4 x\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$

Finally, we have, ${\cot}^{2} x = {\cos}^{2} \frac{x}{\sin} ^ 2 x = \frac{1 + \cos 2 x}{1 - \cos 2 x} \ldots . . \left(4\right)$

Using, $\left(1\right) , \left(3\right) \mathmr{and} \left(4\right) ,$ we have,

$f \left(\theta\right) = \frac{1}{2} \left(1 - \cos 2 \theta\right) + 6 \left\{\frac{1 + \cos 2 \theta}{1 - \cos 2 \theta}\right\} - \left(3 + 4 \cos 2 \theta + \cos 4 \theta\right)$

We can still go further and express $f \left(\theta\right)$ as a rational function, as follows :

f(theta)={(1-cos2theta)^2+12(1+cos2theta)-2(1-cos2theta)(3+4cos2theta+cos4theta)}/{2(1-cos2theta)

Where, $N r . = 1 - 2 \cos 2 \theta + {\cos}^{2} \left(2 \theta\right) + 12 + 12 \cos 2 \theta - 2 \left(3 + 4 \cos 2 \theta + \cos 4 \theta - 3 \cos 2 \theta - 4 {\cos}^{2} \left(2 \theta\right) - \cos 4 \theta \cos 2 \theta\right)$

$= 13 + 10 \cos 2 \theta + {\cos}^{2} \left(2 \theta\right) - 6 - 2 \cos 2 \theta - 2 \cos 4 \theta + 8 {\cos}^{2} \left(2 \theta\right) + 2 \cos 4 \theta \cos 2 \theta$

$= 7 + 8 \cos 2 \theta - 2 \cos 4 \theta + 9 {\cos}^{2} \left(2 \theta\right) + \cos \left(4 \theta + 2 \theta\right) + \cos \left(4 \theta - 2 \theta\right)$

$= 7 + 8 \cos 2 \theta - 2 \cos 4 \theta + 9 \left\{\frac{1 + \cos 4 \theta}{2}\right\} + \cos 6 \theta + \cos 2 \theta$
$= \frac{1}{2} \left(23 + 18 \cos 2 \theta + 5 \cos 4 \theta + 2 \cos 6 \theta\right)$

$\therefore f \left(\theta\right) = \frac{23 + 18 \cos 2 \theta + 5 \cos 4 \theta + 2 \cos 6 \theta}{4 \left(1 - \cos 2 \theta\right)}$