How do you express $sin^4theta+csc^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-04-04T13:39:36

$S = (1 - cos t)^2.(1 + cos t)^2 + 1/[(1 - cos t)(1 + cos t)$

$sin^4 t = sin^2 t.sin^2 t = (1 - cos ^2 t)(1 - cos^2 t) =$ = (1 - cos t)(1 + cos t)(1 - cos t)(1 + cos t) == (1 - cos t)^2.(1 + cos t)^2#

$csc^2 t = 1/(sin^2 t) = 1/[(1 - cos t)(1 + cos t)]$
Finally:
$S = (1 - cos t)^2.(1 + cos t)^2 + 1/[(1 - cos t)(1 + cos t)]$

How do you express sin^4theta+csc^2theta sin^4theta+csc^2theta in terms of non-exponential trigonometric functions?