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Solve the system:x-y=pi/3 sinx+siny=3^1/2

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sodelete | Student, College Freshman | (Level 1) Honors

Posted May 28, 2011 at 10:25 PM via web

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Solve the system:

x-y=pi/3

sinx+siny=3^1/2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 10, 2013 at 3:27 PM (Answer #3)

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You also may use substitution method, such that:

`x - y = pi/3 => x = pi/3 + y`

Taking sine of both sides yields:

`sin x = sin(pi/3 + y)`

Expanding the right side yields:

`sin x = sin(pi/3)*cos y + sin y*cos(pi/3)`

You need to replace `1/2` for `cos(pi/3)` and `sqrt3/2` ` ` for `sin(pi/3) ` such that:

`sin x = sqrt3/2*cos y + 1/2*sin y`

You need to replace `1/2sin y + sqrt3/2cos y` for `sin x` in the bottom equation, such that:

` 1/2sin y + sqrt3/2cos y + sin y = sqrt3 `

You need to keep to the left side the terms that contain sin y, such that:

`sin y(1/2 + 1) = sqrt3(1 - 1/2cos y)`

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Using the basic formula of trigonometry yields:

`sin y = sqrt(1 - cos^2 y)`

`(3/2)sqrt(1 - cos^2 y) = sqrt3(1 - 1/2cos y)`

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Raising to square both sides yields:

`9/4(1 - cos^2 y) = 3(1 - cos y/2)^2`

`9 - 9cos^2 y = 3(4 - 4cos y + cos^2 y)`

`9 - 9cos^2 y = 12 - 12cos y + 3cos^2 y`

`12cos^2 y - 12cos y + 3 = 0 => 4cos^2 y - 4cos y + 1 = 0`

`(2cos y - 1)^2 =0 => 2cos y = 1 => cos y = 1/2 => y = +-cos^(-1)(1/2) => y = +-pi/3`

`x = pi/3 + p/3 => x = 2pi/3`

`x = -pi/3 + pi/3 => x = 0`

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Hence, evaluating the solutions to the given simultaneous equations yields `x = 2pi/3, y = pi/3` ` ` and `` .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 29, 2011 at 3:43 AM (Answer #2)

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To solve the system means to find the pair of values for x and y that verifies the equations of the system.

We'll re-write the second equation of the system, transforming the sum of like trigonometric functions into a product:

sin a + sin b = 2sin [(a+b)/2]*cos [(a-b)/2]

sinx + siny = 2sin [(x+y)/2]*cos [(x-y)/2]

We'll substitute x - y by pi/3 and we'll get:

sinx + siny = 2sin [(x+y)/2]*cos [(pi/3)/2](1)

But sinx + siny = sqrt 3 (2)

We'll put (1) = (2):

2sin [(x+y)/2]*cos [(pi/3)/2] = sqrt 3

We'll divide by 2:

sin [(x+y)/2]*cos [(pi/6)] = (sqrt 3)/2

We'll substitute cos [(pi/6)] = sqrt3/2

(sqrt3/2)*sin [(x+y)/2] = sqrt3/2

We'll divide by sqrt3/2:

sin [(x+y)/2] = 1

(x+y)/2 = arcsin 1

(x+y)/2 = pi/2

x + y = 2pi/2

x + y = pi (3)

x - y = pi/3 (4)

We'll add (3) + (4):

x + y + x - y = pi + pi/3

We'll combine and eliminate like terms:

2x = 4pi/3

We'll divide by 2:

x = 4pi/6

x = 2pi/3

We'll substitute x in (3):

2pi/3 + y = pi

We'll subtract 2pi/3 both sides:

y = pi - 2pi/3

y = pi/3

The solution of the system is represented by the pair (2pi/3 ; pi/3).

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