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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

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

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

Rewrite

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

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

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

= (sin (t) + cos (t)) (sin (t) - cos (t))

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

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

Rewrite using trig identities

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

Expand

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

(sin (t) + cos (t)) (sin (t) - cos (t))

Apply Difference of Two Squares Formula:

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

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

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

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

Use the Double Angle identity:

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

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

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

Simplify

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

Simplify

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

- (2 cos^{2} (t) - 1) \cdot 1

Multiply:

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{1 - cos ( - x )}{sec ( - x ) - 1} = cos (x)

cos (3 θ) = 4 cos^{3} (θ) - 3 cos (θ)

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

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