Is sin(θ)-cos^2(θ)=sin^3(θ) ?

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

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prove sin(θ)-cos^2(θ)=sin^3(θ)

prove

sin (θ) - cos^{2} (θ) = sin^{3} (θ)

False

Solution steps

sin (θ) - cos^{2} (θ) = sin^{3} (θ)

Show that the two sides are not equal

For

sin (θ) - cos^{2} (θ) = sin^{3} (θ)

plug in

θ = 0

sin (0) - cos^{2} (0) = - 1

sin (0) - cos^{2} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 0 - 1^{2}

Simplify

= - 1

sin^{3} (0) = 0

sin^{3} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0^{3}

Simplify

= 0

The two sides are not equal

\Rightarrow False

1 - tan^{2} (a) = cos (2 a) sec^{2} (a)

\frac{sin ( x ) - cos ( x )}{sin ( x )} = 1 - cot (x)

\frac{( sec ( θ ) - 1 )}{( 1 - cos ( θ ) )} = sec (θ)

cos (0) tan (0) = sin (0)

cos (\frac{π}{6}) = \frac{4}{3} cos^{3} (\frac{π}{6})

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