Is sin^2(θ)+1=-2-cos^2(x) ?

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

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prove sin^2(θ)+1=-2-cos^2(x)

prove

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

False

Solution steps

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

Show that the two sides are not equal

For

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

plug in

θ = 0

For

sin^{2} (θ) + 1 = - 2 - cos^{2} (x)

plug in

x = 1

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

sin^{2} (0) + 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

= 0^{2} + 1

Simplify

= 1

- 2 - cos^{2} (1) = - 2.29192 \dots

- 2 - cos^{2} (1)

Refine to a decimal form

= - 2.29192 \dots

The two sides are not equal

\Rightarrow False

sec (x) \cdot cot (x) - sin (x) = cos (x) \cdot cot (x)

\frac{cos ( θ )}{sec ( θ ) + tan ( θ )} = 1 - sin (θ)

\frac{sec ( θ )}{tan ( θ ) + cot ( θ )} = sin (θ)

\frac{1}{cos ^{2} ( x )} - 2 = 0

sin (θ) = cos^{2} (θ) + cos^{2} (θ) \cdot sin (θ)

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