Is 2(cos(θ-1))^2=cos^4(θ)-sin^4(θ) ?

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

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prove 2(cos(θ-1))^2=cos^4(θ)-sin^4(θ)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

2 cos^{2} (0 - 1) = 0.58385 \dots

2 cos^{2} (0 - 1)

Refine to a decimal form

= 0.58385 \dots

cos^{4} (0) - sin^{4} (0) = 1

cos^{4} (0) - sin^{4} (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 1^{4} - 0^{4}

Simplify

= 1

The two sides are not equal

\Rightarrow False

p ro v e cos (x) = \frac{15}{17}

p ro v e csc (A) - sin (A) = (cos (A)) (cot (A))

p ro v e \frac{cos ^{2} ( x )}{cos ^{2} ( x )} = 1

p ro v e \frac{1 - sec ( x )}{csc ( x )} = cos (x) (cot (x))

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

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