What is the general solution for sin^2(a)=1-cos(2a) ?

The general solution for sin^2(a)=1-cos(2a) is a=2pin,a=pi+2pin

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sin^2(a)=1-cos(2a)

Solution

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

Solution

a = 2 πn, a = π + 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - cos (2 a)

from both sides

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

Rewrite using trig identities

- 1 + cos (2 a) + sin^{2} (a)

Use the Double Angle identity:

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

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

= - 1 + cos (2 a) - (cos (2 a) - cos^{2} (a))

Simplify

- 1 + cos (2 a) - (cos (2 a) - cos^{2} (a)) : - 1 + cos^{2} (a)

- 1 + cos (2 a) - (cos (2 a) - cos^{2} (a))

- (cos (2 a) - cos^{2} (a)) : - cos (2 a) + cos^{2} (a)

- (cos (2 a) - cos^{2} (a))

Distribute parentheses

= - (cos (2 a)) - (- cos^{2} (a))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (2 a) + cos^{2} (a)

= - 1 + cos (2 a) - cos (2 a) + cos^{2} (a)

Add similar elements:

cos (2 a) - cos (2 a) = 0

= - 1 + cos^{2} (a)

= - 1 + cos^{2} (a)

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

Solve by substitution

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

Let:

cos (a) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

- 1 + u^{2} + 1 = 0 + 1

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cos (a)

cos (a) = 1, cos (a) = - 1

cos (a) = 1, cos (a) = - 1

cos (a) = 1 : a = 2 πn

cos (a) = 1

General solutions for

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

General solutions for

cos (a) = - 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}

a = π + 2 πn

a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(a)=1-cos(2a) ?
The general solution for sin^2(a)=1-cos(2a) is a=2pin,a=pi+2pin