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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

- 1 + cos^{2} (0) = 0

- 1 + 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

= - 1 + 1^{2}

Simplify

= 0

The two sides are not equal

\Rightarrow False

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

\frac{cos ( x )}{5} = \frac{1}{5} \cdot cos (x)

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

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

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = - 2 + sec^{2} (x)

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