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

Mar 15, 2017

see below

#### Explanation:

Use the Double Argument Property

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1$---> If we solve for ${\cos}^{2} \theta$ then we have $\textcolor{red}{{\cos}^{2} \theta = \frac{1}{2} \left(\cos 2 \theta + 1\right)}$

$\cos 4 \theta = 2 {\cos}^{2} 2 \theta - 1$

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

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

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

$= 8 {\cos}^{4} \theta - 8 {\cos}^{2} \theta + 2 - 1$

$= 8 {\cos}^{4} \theta - 8 {\cos}^{2} \theta + 1$--> If we put in $\frac{1}{2} \left(\cos 2 \theta + 1\right)$ for ${\cos}^{2} \theta$ we have

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

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

$= 8 {\cos}^{4} \theta - 4 \cos 2 \theta - 4 + 1$

$= 8 {\cos}^{4} \theta - 4 \cos 2 \theta - 3$---> Now solve for ${\cos}^{4} \theta$

color(red)(1/ 8 (cos 4 theta+4cos 2 theta+3)= cos^4 theta

Therefore,

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

$= - {\cos}^{2} \theta - \frac{7}{\cos} ^ 2 \theta - \frac{5}{\cos} ^ 4 \theta$

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

:.color(blue)(f(theta)=-1/2 (cos 2theta+1)-14/(cos 2theta+1)- 40/(cos 4 theta+4cos 2 theta+3)