Is sin^2(θ)(1+cos^2(θ))=1 ?

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

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prove sin^2(θ)(1+cos^2(θ))=1

prove

sin^{2} (θ) (1 + cos^{2} (θ)) = 1

False

Solution steps

sin^{2} (θ) (1 + cos^{2} (θ)) = 1

Show that the two sides are not equal

For

sin^{2} (θ) (1 + cos^{2} (θ)) = 1

plug in

θ = 0

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

sin^{2} (0) (1 + 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 + 1^{2})

Simplify

= 0

The two sides are not equal

\Rightarrow False

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p ro v e 2 sin^{2} (\frac{x}{2}) + 4 cos^{2} (\frac{x}{2}) - 3 = cos (x)

p ro v e tan (2 u) = \frac{( 2 tan ( u ) )}{1 - tan ^{2} ( u )}

p ro v e \frac{1}{- sin ( x )} = sec (x) tan (x)

p ro v e 6 cos (θ) = 6 cos (- θ)

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