How do you express ${sin}^{4} θ - {cos}^{3} (\frac{π}{2} - θ)$ $sin^4theta-cos^3(pi/2-theta )$ in terms of non-exponential trigonometric functions?

Question

How do you express ${sin}^{4} θ - {cos}^{3} (\frac{π}{2} - θ)$ $sin^4theta-cos^3(pi/2-theta )$ in terms of non-exponential trigonometric functions?

1 Answer

Impact of this question

2059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-13T08:40:06

$= \frac{1}{4} (1 - 2 cos 2 θ + \frac{1}{2} (1 + cos 4 θ) - 3 sin θ + sin 3 θ)$ $=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)$

Explanation:

Given
$= {sin}^{4} θ - {cos}^{3} (\frac{π}{2} - θ)$ $=sin^4theta-cos^3(pi/2-theta)$

$= {sin}^{4} θ - {sin}^{3} θ$ $=sin^4theta-sin^3theta$

Now we know
${sin}^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ $sin^2theta =1/2(1-cos2theta)$
and
${sin}^{3} θ = \frac{1}{4} (3 sin θ - sin 3 θ)$ $sin^3theta=1/4(3sintheta-sin3theta)$

Inserting these values the given expression becomes

$= {(\frac{1}{2} (1 - cos 2 θ))}^{2} - \frac{1}{4} (3 sin θ - sin 3 θ)$ $=(1/2(1-cos2theta))^2-1/4(3sintheta-sin3theta)$

$= \frac{1}{4} (1 - 2 cos 2 θ + {cos}^{2} 2 θ - 3 sin θ + sin 3 θ)$ $=1/4(1-2cos2theta+cos^2 2theta -3sintheta+sin3theta)$

$= \frac{1}{4} (1 - 2 cos 2 θ + \frac{1}{2} (1 + cos 4 θ) - 3 sin θ + sin 3 θ)$ $=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)$

Science

Math

Humanities

... and beyond

How do you express ${sin}^{4} θ - {cos}^{3} (\frac{π}{2} - θ)$ $sin^4theta-cos^3(pi/2-theta )$ in terms of non-exponential trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you express sin4θ−cos3(π2−θ)sin^4theta-cos^3(pi/2-theta ) in terms of non-exponential trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you express ${sin}^{4} θ - {cos}^{3} (\frac{π}{2} - θ)$ $sin^4theta-cos^3(pi/2-theta )$ in terms of non-exponential trigonometric functions?