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

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

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1=sin(θ)-cos(θ)

Solution

1 = sin (θ) - cos (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = 2 πn + π

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

1 = sin (θ) - cos (θ)

Switch sides

sin (θ) - cos (θ) = 1

Rewrite using trig identities

sin (θ) - cos (θ)

sin (θ) - cos (θ) = 2 sin (θ - \frac{π}{4})

sin (θ) - cos (θ)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (θ) - sin (\frac{π}{4}) cos (θ))

Use the Angle Sum identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (θ - \frac{π}{4})

= 2 sin (θ - \frac{π}{4})

2 sin (θ - \frac{π}{4}) = 1

Divide both sides by

2

2 sin (θ - \frac{π}{4}) = 1

Divide both sides by

2

\frac{2 sin ( θ - \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ - \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ - \frac{π}{4} )}{2} : sin (θ - \frac{π}{4})

\frac{2 sin ( θ - \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (θ - \frac{π}{4})

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (θ - \frac{π}{4}) = \frac{2}{2}

sin (θ - \frac{π}{4}) = \frac{2}{2}

sin (θ - \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (θ - \frac{π}{4}) = \frac{2}{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}

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

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

Solve

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

θ - \frac{π}{4} = \frac{π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

θ - \frac{π}{4} = \frac{π}{4} + 2 πn

Add

\frac{π}{4}

to both sides

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

Simplify

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

Simplify

θ - \frac{π}{4} + \frac{π}{4} : θ

θ - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= θ

Simplify

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

\frac{π}{4} + 2 πn + \frac{π}{4}

Group like terms

= \frac{π}{4} + \frac{π}{4} + 2 πn

Combine the fractions

\frac{π}{4} + \frac{π}{4} : \frac{π}{2}

Apply rule

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

= \frac{π + π}{4}

Add similar elements:

π + π = 2 π

= \frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Solve

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

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

Move

\frac{π}{4}

to the right side

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

Add

\frac{π}{4}

to both sides

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

Simplify

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

Simplify

θ - \frac{π}{4} + \frac{π}{4} : θ

θ - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= θ

Simplify

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

\frac{3 π}{4} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

\frac{π}{4} + \frac{3 π}{4} : π

Apply rule

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

= \frac{π + 3 π}{4}

Add similar elements:

π + 3 π = 4 π

= \frac{4 π}{4}

Divide the numbers:

\frac{4}{4} = 1

= π

= 2 πn + π

θ = 2 πn + π

θ = 2 πn + π

θ = 2 πn + π

θ = \frac{π}{2} + 2 πn, θ = 2 πn + π

Popular Examples

2-cos(θ)=2sin^2(θ)

2 - cos (θ) = 2 sin^{2} (θ)

tan(2x-pi/6)=1,0<= x<= 2pi

tan (2 x - \frac{π}{6}) = 1, 0 \leq x \leq 2 π

cos(kpi)=-1

cos (kπ) = - 1

sin(y)=1

sin (y) = 1

sin(3θ)=-1,0<= θ<= 2pi

sin (3 θ) = - 1, 0 \leq θ \leq 2 π

Frequently Asked Questions (FAQ)

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