What is the general solution for cos(θ/2)=sqrt((1+3/5)/2) ?

The general solution for cos(θ/2)=sqrt((1+3/5)/2) is θ=2*0.46364…+4pin,θ=4pi-2*0.46364…+4pin

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cos(θ/2)=sqrt((1+3/5)/2)

Solution

cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

Solution

θ = 2 \cdot 0.46364 \dots + 4 πn, θ = 4 π - 2 \cdot 0.46364 \dots + 4 πn

Radians

θ = 0.92729 \dots + 12.56637 \dots n, θ = 11.63907 \dots + 12.56637 \dots n

Solution steps

cos (\frac{θ}{2}) = \frac{1 + \frac{3}{5}}{2}

Simplify

\frac{1 + \frac{3}{5}}{2} : \frac{2 5}{5}

\frac{1 + \frac{3}{5}}{2}

\frac{1 + \frac{3}{5}}{2} = \frac{4}{5}

\frac{1 + \frac{3}{5}}{2}

Join

1 + \frac{3}{5} : \frac{8}{5}

1 + \frac{3}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} + \frac{3}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 + 3}{5}

1 \cdot 5 + 3 = 8

1 \cdot 5 + 3

Multiply the numbers:

1 \cdot 5 = 5

= 5 + 3

Add the numbers:

5 + 3 = 8

= 8

= \frac{8}{5}

= \frac{\frac{8}{5}}{2}

Apply the fraction rule:

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

= \frac{8}{5 \cdot 2}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{8}{10}

Cancel the common factor:

2

= \frac{4}{5}

= \frac{4}{5}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{5}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{5}

Rationalize

\frac{2}{5} : \frac{2 5}{5}

\frac{2}{5}

Multiply by the conjugate

\frac{5}{5}

= \frac{2 5}{5 5}

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= \frac{2 5}{5}

= \frac{2 5}{5}

cos (\frac{θ}{2}) = \frac{2 5}{5}

Apply trig inverse properties

cos (\frac{θ}{2}) = \frac{2 5}{5}

General solutions for

cos (\frac{θ}{2}) = \frac{2 5}{5}

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

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn, \frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn, \frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn

Solve

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn : θ = 2 arccos (\frac{2}{5}) + 4 πn

\frac{θ}{2} = arccos (\frac{2 5}{5}) + 2 πn

Simplify

arccos (\frac{2 5}{5}) + 2 πn : arccos (\frac{2}{5}) + 2 πn

arccos (\frac{2 5}{5}) + 2 πn

\frac{2 5}{5} = \frac{2}{5}

\frac{2 5}{5}

Apply radical rule:

5 = 5^{\frac{1}{2}}

= \frac{2 \cdot 5 ^{\frac{1}{2}}}{5}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{5 ^{\frac{1}{2}}}{5 ^{1}} = \frac{1}{5 ^{1 - \frac{1}{2}}}

= \frac{2}{5 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

= \frac{2}{5 ^{\frac{1}{2}}}

Apply radical rule:

5^{\frac{1}{2}} = 5

= \frac{2}{5}

= arccos (\frac{2}{5}) + 2 πn

\frac{θ}{2} = arccos (\frac{2}{5}) + 2 πn

Multiply both sides by

2

\frac{θ}{2} = arccos (\frac{2}{5}) + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 arccos (\frac{2}{5}) + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 arccos (\frac{2}{5}) + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 arccos (\frac{2}{5}) + 2 \cdot 2 πn : 2 arccos (\frac{2}{5}) + 4 πn

2 arccos (\frac{2}{5}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 2 arccos (\frac{2}{5}) + 4 πn

θ = 2 arccos (\frac{2}{5}) + 4 πn

θ = 2 arccos (\frac{2}{5}) + 4 πn

θ = 2 arccos (\frac{2}{5}) + 4 πn

Solve

\frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn : θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

\frac{θ}{2} = 2 π - arccos (\frac{2 5}{5}) + 2 πn

Simplify

2 π - arccos (\frac{2 5}{5}) + 2 πn : 2 π - arccos (\frac{2}{5}) + 2 πn

2 π - arccos (\frac{2 5}{5}) + 2 πn

\frac{2 5}{5} = \frac{2}{5}

\frac{2 5}{5}

Apply radical rule:

5 = 5^{\frac{1}{2}}

= \frac{2 \cdot 5 ^{\frac{1}{2}}}{5}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{5 ^{\frac{1}{2}}}{5 ^{1}} = \frac{1}{5 ^{1 - \frac{1}{2}}}

= \frac{2}{5 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

= \frac{2}{5 ^{\frac{1}{2}}}

Apply radical rule:

5^{\frac{1}{2}} = 5

= \frac{2}{5}

= 2 π - arccos (\frac{2}{5}) + 2 πn

\frac{θ}{2} = 2 π - arccos (\frac{2}{5}) + 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 π - arccos (\frac{2}{5}) + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 π - 2 arccos (\frac{2}{5}) + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot 2 π - 2 arccos (\frac{2}{5}) + 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 2 π - 2 arccos (\frac{2}{5}) + 2 \cdot 2 πn : 4 π - 2 arccos (\frac{2}{5}) + 4 πn

2 \cdot 2 π - 2 arccos (\frac{2}{5}) + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - 2 arccos (\frac{2}{5}) + 4 πn

θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

θ = 2 arccos (\frac{2}{5}) + 4 πn, θ = 4 π - 2 arccos (\frac{2}{5}) + 4 πn

Show solutions in decimal form

θ = 2 \cdot 0.46364 \dots + 4 πn, θ = 4 π - 2 \cdot 0.46364 \dots + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ/2)=sqrt((1+3/5)/2) ?
The general solution for cos(θ/2)=sqrt((1+3/5)/2) is θ=2*0.46364…+4pin,θ=4pi-2*0.46364…+4pin