What is sin(x+y) if sinx+cosy=1/4 and cos x+siny=1/2 ?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

sin (x + y) = sin x* cos y + cos x * sin y

It is given that sin x + cos y = 1/4 and cos x + sin y = 1/2

sin x + cos y = 1/4

square the two sides

=> (sin x)^2 + (cos y)^2 + 2*sin x*cos y = 1/16 ...(1)

cos x + sin y = 1/2

square both the sides

=> (sin y)^2 + (cos x)^2 + 2*sin y*cos x = 1/4 ...(1)

(1) + (2)

=> (sin x)^2 + (cos y)^2 + 2*sin x*cos y + (sin y)^2 + (cos x)^2 + 2*sin y*cos x = 1/16 + 1/4

use the property (sin a)^2 + (cos x)^2 = 1

=> 2 + 2*sin x*cos y + 2*sin y*cos x = 5/16

=> 2*sin x*cos y + 2*sin y*cos x = 5/16 - 2

=> sin x*cos y + sin y*cos x = 5/32 - 1

=> sin x*cos y + sin y*cos x = -27/32

The value of sin (x + y) = -27/32

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll note the given relations as:

sin x + cos y = 1/4 (1)

cos x + sin y = 1/2 (2)

We'll raise to square (1), both sides:

(sin x + cos y)^2 = 1/16

We'll expand the square:

(sin x)^2 + 2sinx*cosy + (cos y)^2 = 1/16 (3)

We'll raise to square (2), both sides:

(cos x + sin y)^2 = 1/4

We'll expand the square:

(cos x)^2 + 2cos x*sin y + (sin y)^2 = 1/4 (4)

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

(sin x)^2 + 2sinx*cosy + (cos y)^2 + (cos x)^2 + 2cos x*sin y + (sin y)^2 = 1/16 + 1/4

But, from the fundamental formula of trigonometry, we'll get:

(sin ax)^2 + (cos x)^2 = 1

(sin y)^2 + (cos y)^2 = 1

1 + 1 + 2(sinx*cosy + cos x*sin y) = 5/16

We'll subtract 2 both sides:

2(sinx*cosy + cos x*sin y) = 5/16 - 2

2(sinx*cosy + cos x*sin y) = -27/16

We'll divide by 2:

sinx*cosy + cos x*sin y = -27/32

But the sum from the left side represents the expanding of sin(x+y):

sin (x+y) = sinx*cosy + cos x*sin y

The value of the sine of the sum of angles x and y is: sin (x+y) = -27/32

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