Is 2sin^2(θ)-1=sin^2(θ)-cos^2(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 2sin^2(θ)-1=sin^2(θ)-cos^2(θ)

prove

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

True

Solution steps

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

Manipulating right side

sin^{2} (θ) - cos^{2} (θ)

Rewrite using trig identities

sin^{2} (θ) - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ) - (1 - sin^{2} (θ))

Simplify

sin^{2} (θ) - (1 - sin^{2} (θ)) : 2 sin^{2} (θ) - 1

sin^{2} (θ) - (1 - sin^{2} (θ))

- (1 - sin^{2} (θ)) : - 1 + sin^{2} (θ)

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

= sin^{2} (θ) - 1 + sin^{2} (θ)

Simplify

sin^{2} (θ) - 1 + sin^{2} (θ) : 2 sin^{2} (θ) - 1

sin^{2} (θ) - 1 + sin^{2} (θ)

Group like terms

= sin^{2} (θ) + sin^{2} (θ) - 1

Add similar elements:

sin^{2} (θ) + sin^{2} (θ) = 2 sin^{2} (θ)

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

= 2 sin^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

(1 + tan^{2} (x)) cos (x) = sec (x)

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

\frac{tan ( β )}{sec ( β )} = sin (β)

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

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