Find sin (x-y) if sinx = -2/5 and siny = 2/3 X is in quadrant IV, and y is in quadrant I
To find sin(x-y), we can use the trigonometric identity: sin(x-y) = sinx * cosy - cosx * siny.
1. Given that sinx = -2/5, we need to find cosx.
Since x is in quadrant IV, where sine is negative and cosine is positive, and Pythagorean theorem states that sin^2(x) + cos^2(x) = 1, we can find cosx by substituting the given values into the equation and solving for cosx:
sin^2(x) + cos^2(x) = 1
(-2/5)^2 + cos^2(x) = 1
4/25 + cos^2(x) = 1
cos^2(x) = 1 - 4/25
cos^2(x) = 25/25 - 4/25
cos^2(x) = 21/25
cosx = ± √(21/25)
Since x is in quadrant IV, where cosine is positive, we take the positive square root:
cosx = √(21/25)
cosx = √(21)/5
2. Given that siny = 2/3, we have the cosine value of y since y is in quadrant I. Using the Pythagorean theorem, we can find cosy:
sin^2(y) + cos^2(y) = 1
(2/3)^2 + cos^2(y) = 1
4/9 + cos^2(y) = 1
cos^2(y) = 1 - 4/9
cos^2(y) = 9/9 - 4/9
cos^2(y) = 5/9
cosy = √(5/9)
cosy = √(5)/3
3. Now we can substitute the values into the sin(x-y) trigonometric identity:
sin(x-y) = sinx * cosy - cosx * siny
sin(x-y) = (-2/5) * (√(5)/3) - (√(21)/5) * (2/3)
Simplifying further:
sin(x-y) = (-2√(5)/15) - (2√(21)/15)
sin(x-y) = [(-2√(5)/15) - (2√(21)/15)] * (15/15)
sin(x-y) = (-2√(5) - 2√(21)) / 15
Therefore, sin(x-y) = (-2√(5) - 2√(21)) / 15.
To find sin(x-y), we can use the trigonometric identity:
sin(x-y) = sinx*cosy - cosx*siny
Given that sinx = -2/5 and siny = 2/3, we need to find cosx and cosy.
Since x is in quadrant IV, sinx is negative and cosx will be positive. We can use the Pythagorean theorem to find the value of cosx:
cosx = sqrt(1 - sin^2(x)) = sqrt(1 - (-2/5)^2) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5
Since y is in quadrant I, both sin(y) and cos(y) are positive. We can use the Pythagorean theorem to find the value of cosy:
cosy = sqrt(1 - sin^2(y)) = sqrt(1 - (2/3)^2) = sqrt(1 - 4/9) = sqrt(5/9) = sqrt(5)/3
Now we can substitute the values of sinx, cosy, cosx, and siny into the formula for sin(x-y):
sin(x-y) = sinx*cosy - cosx*siny
= (-2/5)(sqrt(5)/3) - (sqrt(21)/5)(2/3)
= -2sqrt(5)/15 - 2sqrt(21)/15
= (-2sqrt(5) - 2sqrt(21)) / 15
Therefore, sin(x-y) is equal to (-2sqrt(5) - 2sqrt(21)) / 15.
sinx = -2/5, x is in IV , so
cosx = √21/5
siny = 2/3 , y is in I , so
cosy = √5/3
sin(x-y) = sinxcosy - cosxsiny
= (-2/5)(√5/3) - √21/5)(2/3)
= ( -2√5 - 2√21)/15