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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (b) - sin^{2} (b)

Rewrite using trig identities

cos^{2} (b) - sin^{2} (b)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

cos^{2} (b) - 1 + cos^{2} (b)

Group like terms

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

Add similar elements:

cos^{2} (b) + cos^{2} (b) = 2 cos^{2} (b)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

- sin (- 1) = sin (1)

cos^{4} (x) + sin^{2} (x) = sin^{2} (+ cos^{2} (x))

1 + \frac{1}{tan ^{2} ( a )} = \frac{1}{sin ^{2} ( a )}

\frac{tan ( θ )}{cot ( θ )} = tan^{2} (θ)

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

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