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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

cos^{2} (0) - sin^{2} (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^{2} - 0^{2}

Simplify

= 1

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

2 cos^{2} (- 1)

Refine to a decimal form

= 0.58385 \dots

The two sides are not equal

\Rightarrow False

tan^{2} (x) = (sec (x) - 1) (sec (x) + 1)

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

tan (x) (cos (x) + cot (x)) = sin (x) + 1

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

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

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