What is the general solution for sin^2(α)=1+cos^2(α) ?

The general solution for sin^2(α)=1+cos^2(α) is α= pi/2+pin

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sin^2(α)=1+cos^2(α)

sin^{2} (α) = 1 + cos^{2} (α)

α = \frac{π}{2} + πn

Degrees

α = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (α) = 1 + cos^{2} (α)

Subtract

1 + cos^{2} (α)

from both sides

sin^{2} (α) - 1 - cos^{2} (α) = 0

Rewrite using trig identities

sin^{2} (α) - 1 - cos^{2} (α)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

- cos^{2} (x) + sin^{2} (x) = - cos (2 x)

= - 1 - cos (2 α)

- 1 - cos (2 α) = 0

Move

1

to the right side

- 1 - cos (2 α) = 0

Add

1

to both sides

- 1 - cos (2 α) + 1 = 0 + 1

Simplify

- cos (2 α) = 1

- cos (2 α) = 1

Divide both sides by

- 1

- cos (2 α) = 1

Divide both sides by

- 1

\frac{- cos ( 2 α )}{- 1} = \frac{1}{- 1}

Simplify

cos (2 α) = - 1

cos (2 α) = - 1

General solutions for

cos (2 α) = - 1

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}

2 α = π + 2 πn

2 α = π + 2 πn

Solve

2 α = π + 2 πn : α = \frac{π}{2} + πn

2 α = π + 2 πn

Divide both sides by

2

2 α = π + 2 πn

Divide both sides by

2

\frac{2 α}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

α = \frac{π}{2} + πn

α = \frac{π}{2} + πn

α = \frac{π}{2} + πn

21 cos (\frac{2 π}{687} t) + 228 = 220

cosh (nπ) + sinh (nπ) = 0

sin (\frac{π}{6} x) = \frac{1}{2}

cos (θ) = \frac{65}{65}

- sin (x) - sin (x + \frac{π}{3}) = 0

What is the general solution for sin^2(α)=1+cos^2(α) ?
The general solution for sin^2(α)=1+cos^2(α) is α= pi/2+pin