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

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve the  system:

x-y=pi/3.........(1)

sinx+siny=3^1/2.........(2)

From (1) we get  x = y+pi/3. We substitute this in (2):

sin (x+pi/3) + sinx = sqrt3.

2sin (x+x+pi/3)/2 cos (pi/3)/2 = sqrt3, as sinA +sin B = 2sin(A+B)/2 cos(A-B)/2.

2sin(x+pi/6) cosPi/6 = sqrt3.

2sin (x+pi/6) = (sqrt3)/cosPi/6 = 2

sin(x+pi/6) = 2/2 = 1.

x+pi/6 = 2npi + Pi/2.

x = 2npi+pi/2-pi/6 = 2npi+pi/3.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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