[Note: This one... is super hard for me :( I really need help. please. I only have 2 hours left and I have no idea what I'm doing!!
Suppose that cosx= -3/5 and x is in the third quadrant, and that siny= 2/9 and y is in the second quadrant. Find the exact values
1). sinx
2). cosy
3). tan(y/2)
4). cos(x+y)
5). tan(x+y)
2 answers
It says that i have to show work for each part. If any need my email so they can show me steps from a pic from their phone, It would really be appreciated
Amber, this is a public site, please do not post your phone number, I will ask one of the monitors to delete your above reply.
As to your question, they really are not that difficult. I suggest you sketch a right-angled triangle in the appropriate quadrant and recall
that
sinØ = y/r or opposite / hypotenuse
cosØ = x/r or adjacent/hypotenuse
tanØ = y/x or opposite/adjacent
so if cosx = -3/5 , the angle is in III
then x = -3, r = 5
x^2 + y^2 = r^2
9 + y^2 = 25
y^2 = 16
y = ± 4, but we are in III so y = -4
so sinx = -4/5
(I hope you realize that x in this case is not the best name to use for the angle, please don't confuse the value of x as a side with the angle x )
so we have cosx = -3/5 and sinx = -4/5 ----> 1).
for the second part: given siny = 2/9 , the angle y is in II
Again, a poor choice of name for the angle, don't confuse it with the side y
make your sketch,
the opposite (y) is 2
the hypotenuse (r) is 9
x^2 + y^2 = r^2
x^2 + 4 = 81
x = ±√77, but we are in II, so x = -√77
then cosy = -√77/9 ------> 2).
also, tanx = sinx/cosx
= (-4/5) / (-3/5) = 4/3
tany = siny/cosy
= (2/9) / -(√77/9 = -2/√77
4). cos(x+y) = cosxcosy - sinxsiny
= (-3/5)(-√77/9) - (-4/5)(2/9)
= (3√77 + 8)45
5). since tan(x+y) = sin(x+y) / cos(x+y)
we will first need sin(x+y)
sin(x+y) = sinxcosy + cosxsiny
= (-4/5)(-√77/9) + (-3/5)(2/9)
= (4√77 - 6)/45
Then tan(x+y) = sin(x+y) / cos(x+y)
= ((4√77 - 6)/45) / ((3√77 + 8)45)
= (4√77 - 6) / (3√77 + 8)
3). is a little tricky
recall:
tan 2A = 2tanA/(1 - tan^2 A)
or
tan y = 2tan(y/2)/(1 - tan^2 (y/2) )
I will let tan y/2 = t for easier typing
then:
-2/√77 = 2t/(1 - t^2)
-2/√77 + 2t^2/√77 = 2t
times √77
2t^2 - 2√77 t - 2 = 0
t^2 - √77t - 1 = 0
using the formula:
t = (√77 ± √(77 + 4) )/2
= (√77 ± √81)/2
= √77 + 9)/2 , since y/2 must be in quad I, so the tangent is positive
I checked all my answers with my calculators, they are all correct
As to your question, they really are not that difficult. I suggest you sketch a right-angled triangle in the appropriate quadrant and recall
that
sinØ = y/r or opposite / hypotenuse
cosØ = x/r or adjacent/hypotenuse
tanØ = y/x or opposite/adjacent
so if cosx = -3/5 , the angle is in III
then x = -3, r = 5
x^2 + y^2 = r^2
9 + y^2 = 25
y^2 = 16
y = ± 4, but we are in III so y = -4
so sinx = -4/5
(I hope you realize that x in this case is not the best name to use for the angle, please don't confuse the value of x as a side with the angle x )
so we have cosx = -3/5 and sinx = -4/5 ----> 1).
for the second part: given siny = 2/9 , the angle y is in II
Again, a poor choice of name for the angle, don't confuse it with the side y
make your sketch,
the opposite (y) is 2
the hypotenuse (r) is 9
x^2 + y^2 = r^2
x^2 + 4 = 81
x = ±√77, but we are in II, so x = -√77
then cosy = -√77/9 ------> 2).
also, tanx = sinx/cosx
= (-4/5) / (-3/5) = 4/3
tany = siny/cosy
= (2/9) / -(√77/9 = -2/√77
4). cos(x+y) = cosxcosy - sinxsiny
= (-3/5)(-√77/9) - (-4/5)(2/9)
= (3√77 + 8)45
5). since tan(x+y) = sin(x+y) / cos(x+y)
we will first need sin(x+y)
sin(x+y) = sinxcosy + cosxsiny
= (-4/5)(-√77/9) + (-3/5)(2/9)
= (4√77 - 6)/45
Then tan(x+y) = sin(x+y) / cos(x+y)
= ((4√77 - 6)/45) / ((3√77 + 8)45)
= (4√77 - 6) / (3√77 + 8)
3). is a little tricky
recall:
tan 2A = 2tanA/(1 - tan^2 A)
or
tan y = 2tan(y/2)/(1 - tan^2 (y/2) )
I will let tan y/2 = t for easier typing
then:
-2/√77 = 2t/(1 - t^2)
-2/√77 + 2t^2/√77 = 2t
times √77
2t^2 - 2√77 t - 2 = 0
t^2 - √77t - 1 = 0
using the formula:
t = (√77 ± √(77 + 4) )/2
= (√77 ± √81)/2
= √77 + 9)/2 , since y/2 must be in quad I, so the tangent is positive
I checked all my answers with my calculators, they are all correct
1 answer