Is sin^4(θ)-cos^4(θ)=2sin^2(θ)-1 ?

The answer to whether sin^4(θ)-cos^4(θ)=2sin^2(θ)-1 is True

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prove sin^4(θ)-cos^4(θ)=2sin^2(θ)-1

prove

sin^{4} (θ) - cos^{4} (θ) = 2 sin^{2} (θ) - 1

True

Solution steps

sin^{4} (θ) - cos^{4} (θ) = 2 sin^{2} (θ) - 1

Manipulating left side

sin^{4} (θ) - cos^{4} (θ)

Factor

- cos^{4} (θ) + sin^{4} (θ) : (sin^{2} (θ) + cos^{2} (θ)) (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

- cos^{4} (θ) + sin^{4} (θ)

Rewrite

sin^{4} (θ) - cos^{4} (θ)

(sin^{2} (θ))^{2} - (cos^{2} (θ))^{2}

sin^{4} (θ) - cos^{4} (θ)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (θ) = (sin^{2} (θ))^{2}

= (sin^{2} (θ))^{2} - cos^{4} (θ)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (θ) = (cos^{2} (θ))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

sin^{2} (θ) - cos^{2} (θ)

Apply Difference of Two Squares Formula:

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

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

= (sin (θ) + cos (θ)) (sin (θ) - cos (θ))

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

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

Rewrite using trig identities

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

Expand

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

(sin (θ) + cos (θ)) (sin (θ) - cos (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = sin (θ), b = cos (θ)

= sin^{2} (θ) - cos^{2} (θ)

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

Use the Double Angle identity:

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

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

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

Simplify

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

= - (1 - 2 sin^{2} (θ)) \cdot 1

Simplify

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

- (1 - 2 sin^{2} (θ)) \cdot 1

Multiply:

(1 - 2 sin^{2} (θ)) \cdot 1 = (1 - 2 sin^{2} (θ))

= - (- 2 sin^{2} (θ) + 1)

Distribute parentheses

= - (1) - (- 2 sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (θ)

= - 1 + 2 sin^{2} (θ)

= - 1 + 2 sin^{2} (θ)

= 2 sin^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

(sec (θ) - tan (θ)) (csc (θ) + 1) = cot (θ)

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

cos (θ) csc (θ) = cot (θ)

sin (x - y) + sin (x + y) = 2 sin (x) cos (y)

cot (- x) sin (x) = - cos (x)

Is sin^4(θ)-cos^4(θ)=2sin^2(θ)-1 ?
The answer to whether sin^4(θ)-cos^4(θ)=2sin^2(θ)-1 is True