What is the general solution for 1-cos(θ)=sin(θ/2) ?

The general solution for 1-cos(θ)=sin(θ/2) is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=4pin,θ=2pi+4pin

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1-cos(θ)=sin(θ/2)

Solution

1 - cos (θ) = sin (\frac{θ}{2})

Solution

θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = 4 πn, θ = 2 π + 4 πn

Degrees

θ = 6 0^{\circ} + 72 0^{\circ} n, θ = 30 0^{\circ} + 72 0^{\circ} n, θ = 0^{\circ} + 72 0^{\circ} n, θ = 36 0^{\circ} + 72 0^{\circ} n

Solution steps

1 - cos (θ) = sin (\frac{θ}{2})

Subtract

sin (\frac{θ}{2})

from both sides

1 - cos (θ) - sin (\frac{θ}{2}) = 0

Let:

u = \frac{θ}{2}

1 - cos (2 u) - sin (u) = 0

Rewrite using trig identities

1 - cos (2 u) - sin (u)

Use the Double Angle identity:

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

= 1 - (1 - 2 sin^{2} (u)) - sin (u)

Simplify

1 - (1 - 2 sin^{2} (u)) - sin (u) : 2 sin^{2} (u) - sin (u)

1 - (1 - 2 sin^{2} (u)) - sin (u)

- (1 - 2 sin^{2} (u)) : - 1 + 2 sin^{2} (u)

- (1 - 2 sin^{2} (u))

Distribute parentheses

= - (1) - (- 2 sin^{2} (u))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (u)

= 1 - 1 + 2 sin^{2} (u) - sin (u)

1 - 1 = 0

= 2 sin^{2} (u) - sin (u)

= 2 sin^{2} (u) - sin (u)

- sin (u) + 2 sin^{2} (u) = 0

Solve by substitution

- sin (u) + 2 sin^{2} (u) = 0

Let:

sin (u) = u

- u + 2 u^{2} = 0

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

- u + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} - u = 0

Solve with the quadratic formula

2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 0

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

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

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

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = sin (u)

sin (u) = \frac{1}{2}, sin (u) = 0

sin (u) = \frac{1}{2}, sin (u) = 0

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = \frac{1}{2}

General solutions for

sin (u) = \frac{1}{2}

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}

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

General solutions for

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

Combine all the solutions

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = 2 πn, u = π + 2 πn

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = \frac{π}{6} + 2 πn : θ = \frac{π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

Multiply fractions:

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

= \frac{5 π 2}{6}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{6}

Cancel the common factor:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

\frac{θ}{2} = 2 πn : θ = 4 πn

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 πn

Simplify

θ = 4 πn

θ = 4 πn

\frac{θ}{2} = π + 2 πn : θ = 2 π + 4 πn

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

Simplify

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = 4 πn, θ = 2 π + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1-cos(θ)=sin(θ/2) ?
The general solution for 1-cos(θ)=sin(θ/2) is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=4pin,θ=2pi+4pin