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

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Answer 1 · 2016-06-13T08:40:06

$=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)$

Given
$=sin^4theta-cos^3(pi/2-theta)$

$=sin^4theta-sin^3theta$

Now we know
$sin^2theta =1/2(1-cos2theta)$
and
$sin^3theta=1/4(3sintheta-sin3theta)$

Inserting these values the given expression becomes

$=(1/2(1-cos2theta))^2-1/4(3sintheta-sin3theta)$

$=1/4(1-2cos2theta+cos^2 2theta -3sintheta+sin3theta)$

$=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)$

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