sin,cos,tan數學題 Help!!! plz

1. sinx=24/25
x=73.74(degree) (corr. to 4 sig. fig)

siny=4/5
y=53.13(degree) (corr. to 4 sig. fig)

∴x+y=126.9(degree)

Value of sin (x+y)=0.8
Value of tan (x+y)=-1.33
x+y lies on quardrant II

2. sinx=4/5
x=53.13(degree) (corr. to 4 sig. fig)

cosy=12/13
y=22.62(degree) (corr. to 4 sig. fig)

∴x+y=75.75(degree)
Value of cos (x+y)=0.246
Value of tan (x+y)=3.94
x+y lies on quardrant I

1. If x is in the first and y is in the second quardrant, sinx = 24/25 and siny = 4/5 , find the exact value of sin (x+y) and tan (x+y), and the quadrant in which x+y lies.

If sin x = 24/25, hypotenuse = 25, opposite side = 24, adjacent side = 7,
Then cos x = 7/25, tan x = 24/7
If sin y = 4/5, hypotenuse = 5, opposite side = 4, adjacent side = -3
Then cos y = -3/5, tan y = 4/-3
sin (x + y) = sinx cos y + cos x sin y = (24/25)(-3/5) + (7/25)(4/5)
sin (x + y) = -72/125+ 28/125
sin (x + y) = -44/125 Answer

tan (x + y) = (tan x + tan y)/(1 – tan x tan y) = (24/7 + -4/3)/(1 – (24/7)(-4/3))
tan (x + y) = (44/21)/(39/7) = (44/21)(7/39)
tan (x + y) = 44/117 Answer

tan (x + y) is +ve and sin (x+y) is -ve, x + y is in third quadrant. (III)

2. If x is in the first and y is in the second quardrant, sinx = 4/5, and cosy = 12/13, and cosy = -12/13, find the exact value of cos (x+y) and tan (x+y) , and the quadrant in which x+y lies.

Case 1: cosy = 12/13
If sin x = 4/5, hypotenuse = 5, opposite side = 4, adjacent side = 3
cos x = 3/5, tan x = 4/3
If cos y = 12/13, hypotenuse = 13, opposite side = 5, adjacent side = 12
sin y = 5/13, tan y = 5/12
cos (x+y) = cos x cos y – sin x sin y = (3/5)(12/13) – (4/5)(5/13) = 36/65 – 4/13
cos (x+y) = 16/65 Answer

tan (x + y) = (tan x + tan y)/(1 – tan x tan y) = [(4/3)+ (5/12)]/[(1 – (4/3)(5/12)]
tan (x + y) =(7/4)/(4/9) =(7/4)(9/4)
tan (x + y) = 63/16 Answer

cos (x + y) is +ve and tan (x+y) is +ve, x + y is in first quadrant. (I)

Case (2) cosy = -12/13
If cos y = -12/13, hypotenuse = 13, opposite side = 5, adjacent side = -12
sin y = 5/13, tan y = 5/-12

cos (x+y) = cos x cos y – sin x sin y = (3/5)(-12/13) – (4/5)(5/13) = -36/65 – 4/13
cos (x+y) = -56/65 Answer

tan (x + y) = (tan x + tan y)/(1 – tan x tan y) = [(4/3)+ (5/-12)]/[(1 – (4/3)(5-/12)]
tan (x + y) =(11/12)/(14/9) =(11/12)(9/14)
tan (x + y) = 33/56 Answer

cos (x + y) is -ve and tan (x+y) is +ve, x + y is in third quadrant. (III)



2012-10-08 07:30:46 補充：
In trigonometry,
tan (angle) = opposite side / adjacent side = y/x
sin (angle) = opposite side / hypotenuse = y/r
cos (angle) = adjacent side / hypotenuse = x/r

x^2 + y^2 = r^2