What is the general solution for sin(10)=2sin(a)cos(a) ?

The general solution for sin(10)=2sin(a)cos(a) is a=5+180n,a=90-5+180n

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sin(10)=2sin(a)cos(a)

Solution

sin (1 0^{\circ}) = 2 sin (a) cos (a)

Solution

a = 5^{\circ} + 18 0^{\circ} n, a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

Radians

a = \frac{π}{36} + πn, a = \frac{π}{2} - \frac{π}{36} + πn

Solution steps

sin (1 0^{\circ}) = 2 sin (a) cos (a)

Subtract

2 sin (a) cos (a)

from both sides

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

Rewrite using trig identities

sin (1 0^{\circ}) - 2 sin (a) cos (a)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= sin (1 0^{\circ}) - sin (2 a)

sin (1 0^{\circ}) - sin (2 a) = 0

Move

sin (1 0^{\circ})

to the right side

sin (1 0^{\circ}) - sin (2 a) = 0

Subtract

sin (1 0^{\circ})

from both sides

sin (1 0^{\circ}) - sin (2 a) - sin (1 0^{\circ}) = 0 - sin (1 0^{\circ})

Simplify

- sin (2 a) = - sin (1 0^{\circ})

- sin (2 a) = - sin (1 0^{\circ})

Divide both sides by

- 1

- sin (2 a) = - sin (1 0^{\circ})

Divide both sides by

- 1

\frac{- sin ( 2 a )}{- 1} = \frac{- sin ( 1 0 ^{\circ} )}{- 1}

Simplify

\frac{- sin ( 2 a )}{- 1} = \frac{- sin ( 1 0 ^{\circ} )}{- 1}

Simplify

\frac{- sin ( 2 a )}{- 1} : sin (2 a)

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin (2 a)

Simplify

\frac{- sin ( 1 0 ^{\circ} )}{- 1} : sin (1 0^{\circ})

\frac{- sin ( 1 0 ^{\circ} )}{- 1}

Apply the fraction rule:

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

= \frac{sin ( 1 0 ^{\circ} )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= sin (1 0^{\circ})

sin (2 a) = sin (1 0^{\circ})

sin (2 a) = sin (1 0^{\circ})

sin (2 a) = sin (1 0^{\circ})

Apply trig inverse properties

sin (2 a) = sin (1 0^{\circ})

General solutions for

sin (2 a) = sin (1 0^{\circ})

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n, 2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n, 2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Solve

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n : a = 5^{\circ} + 18 0^{\circ} n

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Divide both sides by

2

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Divide both sides by

2

\frac{2 a}{2} = \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 a}{2} = \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2} : 5^{\circ} + 18 0^{\circ} n

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} = 5^{\circ}

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2}

arcsin (sin (1 0^{\circ})) = 1 0^{\circ}

arcsin (sin (1 0^{\circ}))

For

- 9 0^{\circ} \leq a \leq 9 0^{\circ}, a rcs in (s in (x)) = x

- 9 0^{\circ} \leq 1 0^{\circ} \leq 9 0^{\circ}

= 1 0^{\circ}

= \frac{1 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{18 \cdot 2}

Multiply the numbers:

18 \cdot 2 = 36

= 5^{\circ}

= 5^{\circ} + \frac{36 0 ^{\circ} n}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 5^{\circ} + 18 0^{\circ} n

a = 5^{\circ} + 18 0^{\circ} n

a = 5^{\circ} + 18 0^{\circ} n

a = 5^{\circ} + 18 0^{\circ} n

Solve

2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n : a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Divide both sides by

2

2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Divide both sides by

2

\frac{2 a}{2} = 9 0^{\circ} - \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 a}{2} = 9 0^{\circ} - \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

9 0^{\circ} - \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2} : 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

9 0^{\circ} - \frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2} = 5^{\circ}

\frac{arcsin ( sin ( 1 0 ^{\circ} ) )}{2}

arcsin (sin (1 0^{\circ})) = 1 0^{\circ}

arcsin (sin (1 0^{\circ}))

For

- 9 0^{\circ} \leq a \leq 9 0^{\circ}, a rcs in (s in (x)) = x

- 9 0^{\circ} \leq 1 0^{\circ} \leq 9 0^{\circ}

= 1 0^{\circ}

= \frac{1 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{18 \cdot 2}

Multiply the numbers:

18 \cdot 2 = 36

= 5^{\circ}

= 9 0^{\circ} - 5^{\circ} + \frac{36 0 ^{\circ} n}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

a = 5^{\circ} + 18 0^{\circ} n, a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

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Frequently Asked Questions (FAQ)

What is the general solution for sin(10)=2sin(a)cos(a) ?
The general solution for sin(10)=2sin(a)cos(a) is a=5+180n,a=90-5+180n