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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

2 sin^{2} (0) - 1 = - 1

2 sin^{2} (0) - 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

= 2 \cdot 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

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

(csc (x) - cot (x)) (sec (x) 1) = tan (x)

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

tan^{3} (x) - tan (x) sec^{2} (x) = tan (- x)

sec (π - x) = - sec (x)

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