Is sin^4(β)-cos^4(β)=2sin^2(β)-1 ?

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

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prove sin^4(β)-cos^4(β)=2sin^2(β)-1

prove

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

True

Solution steps

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

Manipulating left side

sin^{4} (β) - cos^{4} (β)

Factor

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

- cos^{4} (β) + sin^{4} (β)

Rewrite

sin^{4} (β) - cos^{4} (β)

(sin^{2} (β))^{2} - (cos^{2} (β))^{2}

sin^{4} (β) - cos^{4} (β)

Apply exponent rule:

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

sin^{4} (β) = (sin^{2} (β))^{2}

= (sin^{2} (β))^{2} - cos^{4} (β)

Apply exponent rule:

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

cos^{4} (β) = (cos^{2} (β))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

sin^{2} (β) - cos^{2} (β)

Apply Difference of Two Squares Formula:

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

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

= (sin (β) + cos (β)) (sin (β) - cos (β))

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

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

Rewrite using trig identities

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

Expand

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

(sin (β) + cos (β)) (sin (β) - cos (β))

Apply Difference of Two Squares Formula:

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

a = sin (β), b = cos (β)

= sin^{2} (β) - cos^{2} (β)

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

Use the Double Angle identity:

cos^{2} (β) - sin^{2} (β) = cos (2 β)

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

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

Simplify

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

= - (1 - 2 sin^{2} (β)) \cdot 1

Simplify

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

- (1 - 2 sin^{2} (β)) \cdot 1

Multiply:

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

= - (- 2 sin^{2} (β) + 1)

Distribute parentheses

= - (1) - (- 2 sin^{2} (β))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (β)

= - 1 + 2 sin^{2} (β)

= - 1 + 2 sin^{2} (β)

= 2 sin^{2} (β) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sin ( π - x )}{sin ( \frac{π}{2} - x )} = tan (x)

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