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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - 2 sin^{2} (x)

Rewrite using trig identities

1 - 2 sin^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

1 - 2 (1 - cos^{2} (x)) : 2 cos^{2} (x) - 1

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

Expand

- 2 (1 - cos^{2} (x)) : - 2 + 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (x)

= - 2 \cdot 1 - (- 2) cos^{2} (x)

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Subtract the numbers:

1 - 2 = - 1

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

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

\frac{1}{sin ( x ) cos ( x )} = 2 csc (2 x)

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