Is 2sin^2(θ)+sin(θ)-1=0 ?

The answer to whether 2sin^2(θ)+sin(θ)-1=0 is False

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prove 2sin^2(θ)+sin(θ)-1=0

prove

2 sin^{2} (θ) + sin (θ) - 1 = 0

False

Solution steps

2 sin^{2} (θ) + sin (θ) - 1 = 0

Show that the two sides are not equal

For

2 sin^{2} (θ) + sin (θ) - 1 = 0

plug in

θ = 0

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

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

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

Simplify

= - 1

The two sides are not equal

\Rightarrow False

\frac{1}{csc ( t ) - cot ( t )} = csc (t) + cot (t)

sin (\frac{5 π}{2} + x) = cos (x)

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

cos (x) = sin (x + \frac{π}{6}) - sin (x - \frac{π}{6})

tan^{2} (\frac{π}{4}) = tan (\frac{π}{4})