Is sin^4(θ)-2sin^2(θ)=cos^4(θ)-1 ?

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

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prove sin^4(θ)-2sin^2(θ)=cos^4(θ)-1

prove

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

True

Solution steps

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

Manipulating left side

sin^{4} (θ) - 2 sin^{2} (θ)

Factor

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

sin^{4} (θ) - 2 sin^{2} (θ)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{4} (θ) = sin^{2} (θ) sin^{2} (θ)

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

Factor out common term

sin^{2} (θ)

= sin^{2} (θ) (sin^{2} (θ) - 2)

= (- 2 + sin^{2} (θ)) sin^{2} (θ)

Rewrite using trig identities

(- 2 + sin^{2} (θ)) sin^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

(- 1 - cos^{2} (θ)) (1 - cos^{2} (θ))

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = - 1, b = - cos^{2} (θ), c = 1, d = - cos^{2} (θ)

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

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

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

Simplify

- 1 \cdot 1 + 1 \cdot cos^{2} (θ) - 1 \cdot cos^{2} (θ) + cos^{2} (θ) cos^{2} (θ) : - 1 + cos^{4} (θ)

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

Add similar elements:

1 \cdot cos^{2} (θ) - 1 \cdot cos^{2} (θ) = 0

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

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

cos^{2} (θ) cos^{2} (θ) = cos^{4} (θ)

cos^{2} (θ) cos^{2} (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{2 + 2} (θ)

Add the numbers:

2 + 2 = 4

= cos^{4} (θ)

= - 1 + cos^{4} (θ)

= - 1 + cos^{4} (θ)

= - 1 + cos^{4} (θ)

= cos^{4} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

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

tan (x - x) = - tan (x)

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

\frac{( cos ( x ) + 1 )}{sin ^{3} ( x )} = \frac{( csc ( x ) )}{1 - cos ( x )}

cos (x) = \frac{sin ( 2 x )}{2 sin ( x )}

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