What is the general solution for sin(piy)=cos(1)sin(pi/6)-1/2 ?

The general solution for sin(piy)=cos(1)sin(pi/6)-1/2 is y=(-0.23192…)/pi+2n,y=1+(0.23192…)/pi+2n

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Popular Trigonometry >

sin(piy)=cos(1)sin(pi/6)-1/2

Solution

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

Solution

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

Degrees

y = - 4.22975 \dots^{\circ} + 114.59155 \dots^{\circ} n, y = 61.52553 \dots^{\circ} + 114.59155 \dots^{\circ} n

Solution steps

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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}

= \frac{1}{2}

= \frac{1}{2}

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

Apply trig inverse properties

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

General solutions for

sin (π y) = cos (1) \frac{1}{2} - \frac{1}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

Solve

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn : y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

Divide both sides by

π

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn

Divide both sides by

π

\frac{π y}{π} = \frac{arcsin ( cos ( 1 ) \frac{1}{2} - \frac{1}{2} )}{π} + \frac{2 πn}{π}

Simplify

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

Solve

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn : y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

Divide both sides by

π

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

Divide both sides by

π

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

Simplify

\frac{π y}{π} = \frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

Simplify

\frac{π y}{π} : y

\frac{π y}{π}

Cancel the common factor:

π

= y

Simplify

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π} : 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

\frac{π}{π} + \frac{arcsin ( - cos ( 1 ) \frac{1}{2} + \frac{1}{2} )}{π} + \frac{2 πn}{π}

Apply rule

\frac{a}{a} = 1

\frac{π}{π} = 1

= 1 + \frac{arcsin ( \frac{1}{2} - \frac{1}{2} cos ( 1 ) )}{π} + \frac{2 πn}{π}

Cancel

\frac{2 πn}{π} : 2 n

\frac{2 πn}{π}

Cancel the common factor:

π

= 2 n

= 1 + \frac{arcsin ( \frac{1}{2} - \frac{1}{2} cos ( 1 ) )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n, y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

Show solutions in decimal form

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(piy)=cos(1)sin(pi/6)-1/2 ?
The general solution for sin(piy)=cos(1)sin(pi/6)-1/2 is y=(-0.23192…)/pi+2n,y=1+(0.23192…)/pi+2n