Is 2sin^2(x)-cos(2x)=1 ?

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

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prove 2sin^2(x)-cos(2x)=1

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

Simplify

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

= 2 \cdot 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

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

\frac{sin ( 2 x )}{1 - sin ^{2} ( x )} = 2 tan (x)

cos (θ) = cos (θ + 2 π)

sin (θ) \cdot cot (θ) = cos (θ)

\frac{1 + sec ( a )}{tan ( a ) + sin ( a )} = csc (a)