Is sin^2(θ)+sin(θ)=1 ?

The answer to whether sin^2(θ)+sin(θ)=1 is False

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prove sin^2(θ)+sin(θ)=1

prove

sin^{2} (θ) + sin (θ) = 1

False

Solution steps

sin^{2} (θ) + sin (θ) = 1

Show that the two sides are not equal

For

sin^{2} (θ) + sin (θ) = 1

plug in

θ = 0

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

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

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

\frac{1}{cos ^{2} ( x )} = 2

sin (θ - \frac{3 π}{2}) = cos (θ)

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

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