Prove the following idenitity cos (t) - cos3 (3 as an exponent) (t)/sin (t) =cos (t) sin (t)?

Question

Please try to explain this answer to me as detailed as possible thanks!

Answer 1

basically: cos(t)-cos^3(t)/sin(t) = cos(t)sin(t). Multiply both sides by sin(t) and divide both sides by cos(t) to achieve:

(cos(t)-cos^3(t))/cos(t) = sin^2(t)

factor out a cos(t) from the numerator or the first expression:

cos(t)(1-cos^2(t))/cos(t) = sin^2(t)

canceling cos(t) gives us:

1-cos^2(t) = sin^2(t)

Rearrange to get:

sin^2(t) + cos^2(t) = 1, which is a true identity. Therefore the first equality is true.

Answer 2

(cos t - cos^3 t)/ sin t = sin t cos t
cos t - cos^3 t = sin^2 t cos t
(cos t - cos^3 t)/ cos t = sin^2 t
(cos t / cos t) - (cos^3 t / cos t) = sin^2 t
1 - cos^2 t = sin ^2 t
sin ^2 t = sin ^2 t

Answer 3

cos (t) - cos^3 (t)
= cos (t) ( 1 - cos^2 (t))
= cos (t) * sin^2 (t)

=> [ cos (t) - cos^3 (t) ] / sin (t)
= [ cos (t) * sin^2 (t) ] / sin (t)
= cos (t) * sin (t)

Answer 4

(cos(t) - cos^3(t))/sin(t) = (cos(t)(1-cos^2(t))/sin(t)
= cos(t)sin^2(t)/sin(t)
= cos(t)sin(t)
= rhs

Answer 5

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