Is sin^2(θ)-cos^2(θ)=1 ?

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

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prove sin^2(θ)-cos^2(θ)=1

prove

sin^{2} (θ) - cos^{2} (θ) = 1

False

Solution steps

sin^{2} (θ) - cos^{2} (θ) = 1

Show that the two sides are not equal

For

sin^{2} (θ) - cos^{2} (θ) = 1

plug in

θ = 0

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

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

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

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

= 0^{2} - 1^{2}

Simplify

= - 1

The two sides are not equal

\Rightarrow False

tan^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

(1 + tan (x)) (1 - cot (x)) = tan (x) - cot (x)

(tan (x) + cot (x))^{2} = sec^{4} (x) cot^{2} (x)

(1 + tan^{2} (s)) (1 - sin^{2} (s)) = 1

8 - 4 csc^{2} (z) = 4 - 4 cot^{2} (z)