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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - 2 sin^{2} (t)

Rewrite using trig identities

1 - 2 sin^{2} (t)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Subtract the numbers:

1 - 2 = - 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

- cos (2 t) sin (2 t) + sin (2 t) cos (2 t) + 0 = 0

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

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

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

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