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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

2 cos (2 \cdot 0) = 2

2 cos (2 \cdot 0)

Simplify

= 2 cos (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

Simplify

= 2

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

2 sin (0^{2}) - 1

Rewrite using trig identities:

sin (0^{2}) = 0

sin (0^{2})

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 \cdot 0 - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

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