Is prove-sin^2(θ)=cos^2(θ) ?

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

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prove-sin^2(θ)=cos^2(θ)

prove

- sin^{2} (θ) = cos^{2} (θ)

False

Solution steps

- sin^{2} (θ) = cos^{2} (θ)

Show that the two sides are not equal

For

- sin^{2} (θ) = cos^{2} (θ)

plug in

θ = 0

- sin^{2} (0) = 0

- sin^{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

= - 0^{2}

Simplify

= 0

cos^{2} (0) = 1

cos^{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

= 1^{2}

Simplify

= 1

The two sides are not equal

\Rightarrow False

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

sec^{2} (x) cot^{2} (x) = cot^{2} (x) + 1

1 + tan^{2} (B) = sec^{2} (B)

sec (\frac{3 π}{4}) = - 2

cos^{2} (7 θ) - sin^{2} (7 θ) = cos (14 θ)

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