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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

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

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

Rewrite

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

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

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Use the Pythagorean identity:

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

= (cos (x) + sin (x)) (cos (x) - sin (x)) \cdot 1

Simplify

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

Expand

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

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

Apply Difference of Two Squares Formula:

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

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

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

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (x) (cot (x) + tan (x)) = sec^{2} (x)

p ro v e \frac{1 + sin ( 2 x )}{sin ( 2 x )} = 1 + \frac{1}{2} sec (x) csc (x)

p ro v e \frac{1}{csc ( x ) + 1} + \frac{1}{sin ( x ) + 1} = 1

p ro v e cos (2 u) = cos^{2} (u) - sin^{2} (u)

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

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