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solvefor α,cos^2(α)+cos^2(β)=1

Solution

solve for

α, cos^{2} (α) + cos^{2} (β) = 1

Solution

α = arccos (- sin (β)) + 2 πn, α = - arccos (- sin (β)) + 2 πn, α = arccos (sin (β)) + 2 πn, α = - arccos (sin (β)) + 2 πn

Solution steps

cos^{2} (α) + cos^{2} (β) = 1

Subtract

1

from both sides

cos^{2} (α) + cos^{2} (β) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (β) + cos^{2} (α)

Use the Pythagorean identity:

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

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

= cos^{2} (α) - sin^{2} (β)

cos^{2} (α) - sin^{2} (β) = 0

Factor

cos^{2} (α) - sin^{2} (β) : (cos (α) + sin (β)) (cos (α) - sin (β))

cos^{2} (α) - sin^{2} (β)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (α) - sin^{2} (β) = (cos (α) + sin (β)) (cos (α) - sin (β))

= (cos (α) + sin (β)) (cos (α) - sin (β))

(cos (α) + sin (β)) (cos (α) - sin (β)) = 0

Solving each part separately

cos (α) + sin (β) = 0 or cos (α) - sin (β) = 0

cos (α) + sin (β) = 0 : α = arccos (- sin (β)) + 2 πn, α = - arccos (- sin (β)) + 2 πn

cos (α) + sin (β) = 0

Move

sin (β)

to the right side

cos (α) + sin (β) = 0

Subtract

sin (β)

from both sides

cos (α) + sin (β) - sin (β) = 0 - sin (β)

Simplify

cos (α) = - sin (β)

cos (α) = - sin (β)

Apply trig inverse properties

cos (α) = - sin (β)

General solutions for

cos (α) = - sin (β)

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

α = arccos (- sin (β)) + 2 πn, α = - arccos (- sin (β)) + 2 πn

α = arccos (- sin (β)) + 2 πn, α = - arccos (- sin (β)) + 2 πn

cos (α) - sin (β) = 0 : α = arccos (sin (β)) + 2 πn, α = - arccos (sin (β)) + 2 πn

cos (α) - sin (β) = 0

Move

sin (β)

to the right side

cos (α) - sin (β) = 0

Add

sin (β)

to both sides

cos (α) - sin (β) + sin (β) = 0 + sin (β)

Simplify

cos (α) = sin (β)

cos (α) = sin (β)

Apply trig inverse properties

cos (α) = sin (β)

General solutions for

cos (α) = sin (β)

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

α = arccos (sin (β)) + 2 πn, α = - arccos (sin (β)) + 2 πn

α = arccos (sin (β)) + 2 πn, α = - arccos (sin (β)) + 2 πn

Combine all the solutions

α = arccos (- sin (β)) + 2 πn, α = - arccos (- sin (β)) + 2 πn, α = arccos (sin (β)) + 2 πn, α = - arccos (sin (β)) + 2 πn

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