Is 2sin^2(θ)-cos(2θ)=1 ?

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

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prove 2sin^2(θ)-cos(2θ)=1

prove

2 sin^{2} (θ) - cos (2 θ) = 1

False

Solution steps

2 sin^{2} (θ) - cos (2 θ) = 1

Show that the two sides are not equal

For

2 sin^{2} (θ) - cos (2 θ) = 1

plug in

θ = 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{1}{csc ( + 1 )} + \frac{1}{sin ( x ) + 1} = 1

\frac{( tan ^{2} ( t ) )}{sec ( t )} = sec (t) - cos (t)

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

3 sin (2 x) = 6 sin (x) cos (x)

\frac{tan ( x )}{csc ( x )} + cos (x) = sec (x)