Is sin(2θ)=2sin^2(θ)-1 ?

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

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prove sin(2θ)=2sin^2(θ)-1

prove

sin (2 θ) = 2 sin^{2} (θ) - 1

False

Solution steps

sin (2 θ) = 2 sin^{2} (θ) - 1

Show that the two sides are not equal

For

sin (2 θ) = 2 sin^{2} (θ) - 1

plug in

θ = 0

sin (2 \cdot 0) = 0

sin (2 \cdot 0)

Simplify

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

2 sin^{2} (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} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

sec (x) - sin (x) = cot (x) cos (x)

\frac{csc ( A )}{csc ( A ) - sin ( A )} = sec^{2} (A)

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

sec (x) = sin (x) tan (x) = cos (x)

tan (θ) = \frac{1}{3}