if sinx=3/5,cosy=12/13,where x and y both lie in 2nd quadrant, find the value of sin(x+y),cos(x-y),tan(x+y)?
To find the value of sin(x+y), we can use the sum formula for sine:
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
Given that sin(x) = 3/5 and cos(y) = 12/13, we need to find cos(x) and sin(y).
Since x lies in the 2nd quadrant, we can use the Pythagorean identity to find cos(x):
sin(x)^2 + cos(x)^2 = 1
(3/5)^2 + cos(x)^2 = 1
9/25 + cos(x)^2 = 1
cos(x)^2 = 1 - 9/25 = 16/25
cos(x) = ±√(16/25) = ±4/5
But since x lies in the 2nd quadrant, cos(x) is negative:
cos(x) = -4/5
Now, let's find sin(y). Since y lies in the 2nd quadrant, we can use the Pythagorean identity again:
sin(y)^2 + cos(y)^2 = 1
sin(y)^2 + (12/13)^2 = 1
sin(y)^2 = 1 - (12/13)^2
sin(y) = ±√[1 - (12/13)^2] = ±√(1 - 144/169) = ±√(169/169 - 144/169)
sin(y) = ±√(25/169) = ±5/13
But since y lies in the 2nd quadrant, sin(y) is positive:
sin(y) = 5/13
Now we have all the values needed.
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
= (3/5)(12/13) + (-4/5)(5/13)
= 36/65 - 20/65
= 16/65
cos(x-y) = cos(x)cos(y) + sin(x)sin(y)
= (-4/5)(12/13) + (3/5)(5/13)
= -48/65 + 15/65
= -33/65
tan(x+y) = sin(x+y) / cos(x+y)
= (16/65) / (-33/65)
= -16/33
= -8/11
Therefore, the value of sin(x+y) is 16/65, cos(x-y) is -33/65, and tan(x+y) is -8/11.
To solve this problem, we can use the trigonometric identities to find the values of sin(x+y), cos(x-y), and tan(x+y). The given information is sin(x) = 3/5 and cos(y) = 12/13, where x and y both lie in the second quadrant.
First, let's find the values of sin(x) and cos(x):
Since sin(x) = 3/5 and x lies in the second quadrant, sin(x) is positive, but cos(x) is negative.
Using the Pythagorean identity, sin^2(x) + cos^2(x) = 1, we find:
(3/5)^2 + cos^2(x) = 1
9/25 + cos^2(x) = 1
cos^2(x) = 16/25
cos(x) = -4/5 [taking the negative root since cos(x) is negative in the second quadrant]
Next, let's find the values of sin(y) and cos(y):
Given that cos(y) = 12/13 and y lies in the second quadrant, sin(y) is positive.
Using the Pythagorean identity, sin^2(y) + cos^2(y) = 1, we find:
sin^2(y) + (12/13)^2 = 1
sin^2(y) = 169/169 - 144/169
sin^2(y) = 25/169
sin(y) = 5/13 [taking the positive root since sin(y) is positive in the second quadrant]
Now, let's compute sin(x+y):
Using the sum formula for sine, sin(x+y) = sin(x)cos(y) + cos(x)sin(y), we can substitute in the values we found earlier:
sin(x+y) = (3/5)(12/13) + (-4/5)(5/13)
sin(x+y) = 36/65 - 20/65
sin(x+y) = 16/65
Next, let's compute cos(x-y):
Using the difference formula for cosine, cos(x-y) = cos(x)cos(y) + sin(x)sin(y), we can substitute in the values we found earlier:
cos(x-y) = (-4/5)(12/13) + (3/5)(5/13)
cos(x-y) = -48/65 + 15/65
cos(x-y) = -33/65
Finally, let's compute tan(x+y):
Using the tangent identity, tan(x+y) = sin(x+y)/cos(x+y), we can substitute in the values we found earlier:
tan(x+y) = (16/65) / (-33/65)
tan(x+y) = -16/33
-56/65
in QII, cosy is negative
cosx = -4/5
siny = 5/13
now, just plug in the formulas. for example
sin(x+y) = sinx*cosy+cosx*siny
= (3/5)(-12/13)+(-4/5)(5/13)
= -36/65 - 20/65
= -56/65