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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

sin^{2} (0) = 0

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

= 0^{2}

Simplify

= 0

cos (2 \cdot 0) - 2 = - 1

cos (2 \cdot 0) - 2

Simplify

= cos (0) - 2

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

= 1 - 2

Simplify

= - 1

The two sides are not equal

\Rightarrow False

\frac{tan ( θ ) cot ( θ )}{cos ( θ )} = sec (θ)

2 (cos (θ - 1))^{2} = cos^{4} (θ) - sin^{4} (θ)

\frac{sec ^{2} ( \frac{x}{2} )}{2} = \frac{1}{1 + cos ( x )}

cos (x) = \frac{15}{17}

tan (b) + cot (b) = \frac{1}{sin ( b ) cos ( b )}