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
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
2 answers
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
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