Is (1-cos(θ))+sin^2(θ)=2 ?

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

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prove (1-cos(θ))+sin^2(θ)=2

prove

(1 - cos (θ)) + sin^{2} (θ) = 2

False

Solution steps

1 - cos (θ) + sin^{2} (θ) = 2

Show that the two sides are not equal

For

1 - cos (θ) + sin^{2} (θ) = 2

plug in

θ = 0

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

1 - cos (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 - 1 + 0^{2}

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e \frac{1}{sec ( θ )} = (sec (θ))^{- 1}

p ro v e sin (2 θ) + cos (2 θ) = 1

p ro v e 25 (sec^{2} (5 x) - tan^{2} (5 x)) = 25

p ro v e cot^{2} (x) = \frac{( cos ^{2} ( x ) )}{1 - cos ^{2} ( x )}

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