Is sin^2(θ)=1+cos^2(θ) ?

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

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prove sin^2(θ)=1+cos^2(θ)

prove

sin^{2} (θ) = 1 + cos^{2} (θ)

False

Solution steps

sin^{2} (θ) = 1 + cos^{2} (θ)

Show that the two sides are not equal

For

sin^{2} (θ) = 1 + 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

1 + cos^{2} (0) = 2

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 + 1^{2}

Simplify

= 2

The two sides are not equal

\Rightarrow False

sin (\frac{π}{2}) = \frac{4}{2}

sec (\frac{π}{4} + a) sec (\frac{π}{4} - a) = 2 sec (2 a)

cot^{2} (\frac{v}{2}) = \frac{( sec ( v ) + 1 )}{( sec ( v ) - 1 )}

cot (3) = \frac{cos ( 3 )}{sin ( 3 )}

cos (2 θ) - cos (θ) + 1 = 0