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

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

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Popular >

sin(θ)cos(θ)=-1/2

Solution

sin (θ) cos (θ) = - \frac{1}{2}

Solution

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (θ) cos (θ) = - \frac{1}{2}

Rewrite using trig identities

sin (θ) cos (θ)

Use the Double Angle identity:

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

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 θ )}{2}

\frac{sin ( 2 θ )}{2} = - \frac{1}{2}

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = - \frac{1}{2}

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 sin ( 2 θ )}{2} : sin (2 θ)

\frac{2 sin ( 2 θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (2 θ)

Simplify

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

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

Remove parentheses:

(- a) = - a

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

Multiply fractions:

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

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

Cancel the common factor:

2

= - 1

sin (2 θ) = - 1

sin (2 θ) = - 1

sin (2 θ) = - 1

General solutions for

sin (2 θ) = - 1

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}

2 θ = \frac{3 π}{2} + 2 πn

2 θ = \frac{3 π}{2} + 2 πn

Solve

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

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Popular Examples

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cos(x)=((sqrt(3))/2)

cos (x) = (\frac{3}{2})

csc^2(θ)-9csc(θ)+20=0

csc^{2} (θ) - 9 csc (θ) + 20 = 0

Frequently Asked Questions (FAQ)

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