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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Expand

(- 1 + sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

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

Apply Perfect Square Formula:

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

a = - 1, b = sin^{2} (x)

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

Simplify

(- 1)^{2} + 2 (- 1) sin^{2} (x) + (sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

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

Remove parentheses:

(- a) = - a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

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

Apply exponent rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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