What is ${sec}^{4} θ$ $sec^4theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-06-15T14:00:37

$= \frac{8}{3 + 4 cos 2 θ + cos 4 θ}$ $=8/(3+4cos2theta+cos4theta)$

${sec}^{4} θ = \frac{1}{{cos}^{4} θ}$ $sec^4theta=1/cos^4theta$

$= \frac{1}{{(\frac{1}{2} (1 + cos 2 θ))}^{2}}$ $=1/(1/2(1+cos2theta))^2$

$= \frac{1}{\frac{1}{4} (1 + 2 cos 2 θ + {cos}^{2} 2 θ)}$ $=1/(1/4(1+2cos2theta+cos^2 2theta)$

$= \frac{4}{1 + 2 cos 2 θ + \frac{1}{2} (1 + cos 4 θ)}$ $=4/(1+2cos2theta+1/2(1+cos4theta))$

$= \frac{8}{3 + 4 cos 2 θ + cos 4 θ}$ $=8/(3+4cos2theta+cos4theta)$

What is sec4θsec^4theta in terms of non-exponential trigonometric functions?