Given:
sin x = 4/5, 0 < x < π/2
sin y = 5/13, π/2 < y < π
Find the exact value of sin(x + y)
I presume I'm supposed to use the sum and difference formulas but I'm not sure how to get the exactly value of cos x or cos y
make sketches of each triangle
for sinx = 4/5
sin^2 + cos^2 x = 1
16/25 + cos^2 x = 1
cos^2 x = 9/25
cos x = 3/5, in quadrant I
siny = 5/13
using your triangle and Pythagoras
cosy = -12/13 , since we are in quadrant II
sin(x+y)
= sinxcosy + cosxsiny
= (4/5)(-12/13) + (3/5)(5/13)
= -33/65
cos = √(1 - sin^2)
0 + π/2 < x + y < π/2 + π
... π/2 < x + y < 3π/2
the sign of the solution will be neg
To find the exact value of sin(x + y), we can use the sum formula for sine:
sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
First, let's find cos(x) and cos(y) by using the Pythagorean identity:
cos^2(x) = 1 - sin^2(x),
cos^2(y) = 1 - sin^2(y).
Given sin(x) = 4/5, we can find cos(x) as:
cos^2(x) = 1 - (4/5)^2,
cos^2(x) = 1 - 16/25,
cos^2(x) = 9/25,
cos(x) = ±√(9/25),
cos(x) = ±(3/5).
Given sin(y) = 5/13, we can find cos(y) as:
cos^2(y) = 1 - (5/13)^2,
cos^2(y) = 1 - 25/169,
cos^2(y) = 144/169,
cos(y) = ±√(144/169),
cos(y) = ±(12/13).
Since x lies in the first quadrant (0 < x < π/2), the cosine values of x and y should be positive.
Therefore, cos(x) = 3/5, and cos(y) = 12/13.
Now, let's substitute the values into the sum formula:
sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
= (4/5) * (12/13) + (3/5) * (5/13)
= (48/65) + (15/65)
= 63/65.
So, the exact value of sin(x + y) is 63/65.