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

Question

1347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T09:54:20

$f(θ)=−sin^2(θ)−2cos^2(θ)−5sin^4θ$

$cos2theta = cos^2theta - sin^2theta$
$cos2theta= 2cos^2theta - 1$
$cos2theta = 1- 2sin^2theta$

$=>(cos2theta-1)/2−(cos2theta+1)−5(sin^2θ)^2$

$=>(cos2theta-1)/2−(cos2theta+1)−5((1-cos2theta)/2)^2$

$=>(cos2theta-1)/2−(cos2theta+1)−5/4(1-cos2theta)^2$

$=>(cos2theta-1)/2−(cos2theta+1)−5/4(1+cos^2 2theta - 2cos2theta)$

$=>(cos2theta-1)/2−(cos2theta+1)−5/4(1+(cos4theta+1)/2 - 2cos2theta)$

$=>(2cos2theta-2)/4−(4cos2theta+4)/4−5/4(1+(cos4theta+1)/2 - 2cos2theta)$

$=>1/4(2cos2theta-2−4cos2theta-4−5-(5cos4theta)/2-5/2 + 10cos2theta)$

$=>1/4(8cos2theta−25/2-(5cos4theta)/2-5/2)$

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