Is 1-2cos^2(θ)+cos^4(θ)=sin^4(θ) ?

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

prove 1-2cos^2(θ)+cos^4(θ)=sin^4(θ)

prove

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

True

Solution steps

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

Manipulating left side

1 - 2 cos^{2} (θ) + cos^{4} (θ)

Apply Perfect Square Formula:

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

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

= (1 - cos^{2} (θ))^{2}

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

= (sin^{2} (θ))^{2}

Simplify

(sin^{2} (θ))^{2} : sin^{4} (θ)

(sin^{2} (θ))^{2}

Apply exponent rule:

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

= sin^{2 \cdot 2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (θ)

= sin^{4} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e csc^{2} (θ) sec^{2} (θ) - csc^{2} (θ) = sec^{2} (θ)

p ro v e tan (a) + cot (a) = \frac{csc ( a )}{cos ( a )}

p ro v e sin (2 x) csc (2 x) = 1

p ro v e \frac{cot ^{2} ( x ) - 1}{csc ( x ) + 1} = csc (x) - 1

p ro v e \frac{( cos ( π + x ) )}{cos ( \frac{3 π}{2} - x )} = cot (x)

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