What is the general solution for cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi ?

Q: What is the general solution for cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi ?

The general solution for cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3

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Popular >

cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi

Solution

cos (θ) + sin (θ + \frac{π}{2}) = 1, 0 \leq θ < 2 π

Solution

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Solution steps

cos (θ) + sin (θ + \frac{π}{2}) = 1, 0 \leq θ < 2 π

Rewrite using trig identities

cos (θ) + sin (θ + \frac{π}{2}) = 1

Rewrite using trig identities

sin (θ + \frac{π}{2})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

Simplify

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

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

sin (θ) cos (\frac{π}{2}) = 0

sin (θ) cos (\frac{π}{2})

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) sin (\frac{π}{2}) = cos (θ)

cos (θ) sin (\frac{π}{2})

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:

cos (θ) \cdot 1 = cos (θ)

= cos (θ)

= 0 + cos (θ)

0 + cos (θ) = cos (θ)

= cos (θ)

= cos (θ)

cos (θ) + cos (θ) = 1

Simplify

cos (θ) + cos (θ) : 2 cos (θ)

cos (θ) + cos (θ)

Add similar elements:

cos (θ) + cos (θ) = 2 cos (θ)

= 2 cos (θ)

2 cos (θ) = 1

2 cos (θ) = 1

Subtract

1

from both sides

2 cos (θ) - 1 = 0

Move

1

to the right side

2 cos (θ) - 1 = 0

Add

1

to both sides

2 cos (θ) - 1 + 1 = 0 + 1

Simplify

2 cos (θ) = 1

2 cos (θ) = 1

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

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}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

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cos(θ)= 2/3 ,-(5pi)/2 ,2pi

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi ?
The general solution for cos(θ)+sin(θ+pi/2)=1,0<= θ<2pi is θ= pi/3 ,θ=(5pi)/3