Is sin^3(θ)+sin(θ)cos^2(θ)=sin(θ) ?

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

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prove sin^3(θ)+sin(θ)cos^2(θ)=sin(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin^{3} (θ) + sin (θ) cos^{2} (θ)

Rewrite using trig identities

sin^{3} (θ) + sin (θ) cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{3} (θ) + sin (θ) (1 - sin^{2} (θ))

Simplify

sin^{3} (θ) + sin (θ) (1 - sin^{2} (θ)) : sin (θ)

sin^{3} (θ) + sin (θ) (1 - sin^{2} (θ))

Expand

sin (θ) (1 - sin^{2} (θ)) : sin (θ) - sin^{3} (θ)

sin (θ) (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = sin (θ), b = 1, c = sin^{2} (θ)

= sin (θ) \cdot 1 - sin (θ) sin^{2} (θ)

= 1 \cdot sin (θ) - sin^{2} (θ) sin (θ)

Simplify

1 \cdot sin (θ) - sin^{2} (θ) sin (θ) : sin (θ) - sin^{3} (θ)

1 \cdot sin (θ) - sin^{2} (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

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 (θ) - sin^{3} (θ)

= sin (θ) - sin^{3} (θ)

= sin^{3} (θ) + sin (θ) - sin^{3} (θ)

Add similar elements:

sin^{3} (θ) - sin^{3} (θ) = 0

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

sec^{2} (θ) cot^{2} (θ) = csc^{2} (θ)

tan (\frac{3 π}{2} - x) = cot (x)

\frac{3 csc ( - x )}{sec ( - x )} = - 3 cot (x)

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

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

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