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

sin (x + \frac{π}{2}) = cos (x)

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

csc^{4} (x) - 2 csc^{2} (x) + 1 = cot^{4} (x)

\frac{cos ( x )}{1 - sin ( x )} + \frac{1 - sin ( x )}{cos ( x )} = 2 sec (x)

\frac{tan ( θ ) cot ( θ )}{csc ( θ )} = sin (θ)

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