Is sin(θ)-sin(θ)cos^2(θ)=sin^3(θ) ?

The answer to whether sin(θ)-sin(θ)cos^2(θ)=sin^3(θ) is True

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prove sin(θ)-sin(θ)cos^2(θ)=sin^3(θ)

prove

sin (θ) - sin (θ) cos^{2} (θ) = sin^{3} (θ)

True

Solution steps

sin (θ) - sin (θ) cos^{2} (θ) = sin^{3} (θ)

Manipulating left side

sin (θ) - sin (θ) cos^{2} (θ)

Factor

sin (θ) - cos^{2} (θ) sin (θ) : sin (θ) (1 - cos^{2} (θ))

sin (θ) - cos^{2} (θ) sin (θ)

Factor out common term

sin (θ)

= sin (θ) (1 - cos^{2} (θ))

= (1 - cos^{2} (θ)) sin (θ)

Rewrite using trig identities

(1 - cos^{2} (θ)) sin (θ)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= sin^{2} (θ) sin (θ)

Simplify

sin^{2} (θ) sin (θ) : sin^{3} (θ)

sin^{2} (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= sin^{3} (θ)

= sin^{3} (θ)

= sin^{3} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

p ro v e \frac{tan ( θ ) cot ( θ )}{csc ( θ )} = sin (θ)

p ro v e (1 - cos (x)) (1 + cos (x)) = \frac{1}{csc ^{2} ( x )}

p ro v e csc (x) cos (x) = cot (x)

p ro v e sin (π - x) = sin (x)

Is sin(θ)-sin(θ)cos^2(θ)=sin^3(θ) ?
The answer to whether sin(θ)-sin(θ)cos^2(θ)=sin^3(θ) is True