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

The general solution for 2sin(pi/2-θ)=sin(-θ) is θ=-1.10714…+pin

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Popular Trigonometry >

2sin(pi/2-θ)=sin(-θ)

Solution

2 sin (\frac{π}{2} - θ) = sin (- θ)

Solution

θ = - 1.10714 \dots + πn

Degrees

θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (\frac{π}{2} - θ) = sin (- θ)

Rewrite using trig identities

2 sin (\frac{π}{2} - θ) = sin (- θ)

Rewrite using trig identities

sin (\frac{π}{2} - θ)

Use the Angle Difference identity:

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

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

Simplify

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

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

sin (\frac{π}{2}) cos (θ) = cos (θ)

sin (\frac{π}{2}) cos (θ)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

cos (\frac{π}{2}) sin (θ) = 0

cos (\frac{π}{2}) sin (θ)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0 \cdot sin (θ)

Apply rule

0 \cdot a = 0

= 0

= cos (θ) - 0

cos (θ) - 0 = cos (θ)

= cos (θ)

= cos (θ)

2 cos (θ) = - sin (θ)

2 cos (θ) = - sin (θ)

Subtract

- sin (θ)

from both sides

2 cos (θ) + sin (θ) = 0

Rewrite using trig identities

2 cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{2 cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

2 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

2 + tan (θ) = 0

2 + tan (θ) = 0

Move

2

to the right side

2 + tan (θ) = 0

Subtract

2

from both sides

2 + tan (θ) - 2 = 0 - 2

Simplify

tan (θ) = - 2

tan (θ) = - 2

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

Show solutions in decimal form

θ = - 1.10714 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(pi/2-θ)=sin(-θ) ?
The general solution for 2sin(pi/2-θ)=sin(-θ) is θ=-1.10714…+pin