Is sin^4(x)-cos^4(x)=2sin^2(x)-1 ?

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

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prove sin^4(x)-cos^4(x)=2sin^2(x)-1

prove

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

True

Solution steps

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

Manipulating left side

sin^{4} (x) - cos^{4} (x)

Factor

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

- cos^{4} (x) + sin^{4} (x)

Rewrite

sin^{4} (x) - cos^{4} (x)

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

sin^{4} (x) - cos^{4} (x)

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Rewrite using trig identities

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

Expand

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

(sin (x) + cos (x)) (sin (x) - cos (x))

Apply Difference of Two Squares Formula:

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

a = sin (x), b = cos (x)

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

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

Use the Double Angle identity:

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

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

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

Simplify

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

Simplify

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

- (1 - 2 sin^{2} (x)) \cdot 1

Multiply:

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin^{3} (2 x) = (\frac{1}{2} sin (2 x)) (1 - cos (4 x))

cos (θ) tan (θ) = sin (θ)

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

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