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

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

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Popular >

sin(θ)=sqrt(1+(1-(cos(θ)-1)^2))

Solution

sin (θ) = 1 + (1 - (cos (θ) - 1)^{2})

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = 1 + (1 - (cos (θ) - 1)^{2})

Square both sides

sin^{2} (θ) = (1 + 1 - (cos (θ) - 1)^{2})^{2}

Subtract

1 + 1 - (cos (θ) - 1)^{2}^{2}

from both sides

sin^{2} (θ) - 2 cos (θ) + cos^{2} (θ) - 1 = 0

Rewrite using trig identities

sin^{2} (θ) - 2 cos (θ) + cos^{2} (θ) - 1

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= - 1 - 2 cos (θ) + 1

- 1 - 2 cos (θ) + 1 = 0

Group like terms

- 2 cos (θ) - 1 + 1 = 0

- 1 + 1 = 0

- 2 cos (θ) = 0

Divide both sides by

- 2

- 2 cos (θ) = 0

Divide both sides by

- 2

\frac{- 2 cos ( θ )}{- 2} = \frac{0}{- 2}

Simplify

cos (θ) = 0

cos (θ) = 0

General solutions for

cos (θ) = 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}

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

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (θ) = 1 + 1 - (cos (θ) - 1)^{2}

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sin (θ) = 1 + 1 - (cos (θ) - 1)^{2}

plug in

θ = \frac{π}{2} + 2 π 1

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

Refine

1 = 1

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

sin (θ) = 1 + 1 - (cos (θ) - 1)^{2}

plug in

θ = \frac{3 π}{2} + 2 π 1

sin (\frac{3 π}{2} + 2 π 1) = 1 + 1 - (cos (\frac{3 π}{2} + 2 π 1) - 1)^{2}

Refine

- 1 = 1

\Rightarrow False

θ = \frac{π}{2} + 2 πn

Popular Examples

0=4cos(4x)

0 = 4 cos (4 x)

tan(θ)= 8/7 ,pi<θ<(3pi)/2

tan (θ) = \frac{8}{7}, π < θ < \frac{3 π}{2}

csc(θ)sin(θ)=cot(θ)

csc (θ) sin (θ) = cot (θ)

7sin(θ)+cos(θ)=5

7 sin (θ) + cos (θ) = 5

tan(θ+pi)+2sin(θ+pi)=0

tan (θ + π) + 2 sin (θ + π) = 0

Frequently Asked Questions (FAQ)

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