Is sin(θ)=cos^2(θ)+cos^2(θ)*sin(θ) ?

The answer to whether sin(θ)=cos^2(θ)+cos^2(θ)*sin(θ) is False

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prove sin(θ)=cos^2(θ)+cos^2(θ)*sin(θ)

prove

sin (θ) = cos^{2} (θ) + cos^{2} (θ) \cdot sin (θ)

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

sin (0) = 0

sin (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

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

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

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

= 1^{2} + 1^{2} \cdot 0

Simplify

= 1

The two sides are not equal

\Rightarrow False

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

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

tan (θ) sec (θ) (csc (θ) - sin (θ)) = 1

7 cos^{2} (x) + 5 sin^{2} (x) = 6 + cos (2 x)

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