What is the general solution for sin^2(θ)=2sin^2(θ/2) ?

The general solution for sin^2(θ)=2sin^2(θ/2) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=2pin

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sin^2(θ)=2sin^2(θ/2)

Solution

sin^{2} (θ) = 2 sin^{2} (\frac{θ}{2})

Solution

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (θ) = 2 sin^{2} (\frac{θ}{2})

Subtract

2 sin^{2} (\frac{θ}{2})

from both sides

sin^{2} (θ) - 2 sin^{2} (\frac{θ}{2}) = 0

Rewrite using trig identities

sin^{2} (θ) - 2 sin^{2} (\frac{θ}{2})

Use the Pythagorean identity:

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

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

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

Simplify

1 - cos^{2} (θ) - 2 (1 - cos^{2} (\frac{θ}{2})) : 2 cos^{2} (\frac{θ}{2}) - cos^{2} (θ) - 1

1 - cos^{2} (θ) - 2 (1 - cos^{2} (\frac{θ}{2}))

Expand

- 2 (1 - cos^{2} (\frac{θ}{2})) : - 2 + 2 cos^{2} (\frac{θ}{2})

- 2 (1 - cos^{2} (\frac{θ}{2}))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (\frac{θ}{2})

= - 2 \cdot 1 - (- 2) cos^{2} (\frac{θ}{2})

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (\frac{θ}{2})

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (\frac{θ}{2})

= 1 - cos^{2} (θ) - 2 + 2 cos^{2} (\frac{θ}{2})

Simplify

1 - cos^{2} (θ) - 2 + 2 cos^{2} (\frac{θ}{2}) : 2 cos^{2} (\frac{θ}{2}) - cos^{2} (θ) - 1

1 - cos^{2} (θ) - 2 + 2 cos^{2} (\frac{θ}{2})

Group like terms

= - cos^{2} (θ) + 2 cos^{2} (\frac{θ}{2}) + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

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

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

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

Use the Double Angle identity:

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

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

Multiply

2 \cdot \frac{θ}{2} : θ

2 \cdot \frac{θ}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{θ \cdot 2}{2}

Cancel the common factor:

2

= θ

= - cos^{2} (θ) + cos (θ)

cos (θ) - cos^{2} (θ) = 0

Solve by substitution

cos (θ) - cos^{2} (θ) = 0

Let:

cos (θ) = u

u - u^{2} = 0

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

u - u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 ( - 1 )}, u_{2} = \frac{- 1 - 1}{2 ( - 1 )}

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

\frac{- 1 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 1}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 1 )} : 1

\frac{- 1 - 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 1}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)=2sin^2(θ/2) ?
The general solution for sin^2(θ)=2sin^2(θ/2) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=2pin