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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

1 - 2 sin^{2} (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

= 1 - 2 \cdot 0^{2}

Simplify

= 1

2 cos^{2} (0) = 2

2 cos^{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 1^{2}

Simplify

= 2

The two sides are not equal

\Rightarrow False

cos (θ) = 3

1.5 sin (2 a) cos (2 a) = sin (3 a)

\frac{1}{1 - cot ( x )} = 1 - tan (x)

\frac{5 sin ( 2 x )}{2} = \frac{5}{2} (2 sin (x) cos (x))

\frac{sin ( 2 θ ) + sin ( θ )}{cos ( 2 θ ) + cos ( θ ) + 1} = tan (θ)