if cos(x) = 5/13 and sin(y)=3/5 where 0<= x, y <= x/2, determine:
a) cos (x+y)
b) sin (x-y)
sin x = sqrt (1-cos^2 x) = sqrt (1-25/169)=sqrt(144/169) = 12/13
cos y = sqrt (1-sin^2 y) = sqrt (1-9/25) = sqrt(16/25) = 4/5
cos(x+y) = cos x cos y - sin x sin y
= 5/13 * 4/5 - 12/13 * 3/5
etc
then
sin(x-y) = sin x cos y - cos x sin y
You should recognize the two infamous right-angled triangles with
side ratios of 3-4-5, and 5-12-13
If you don't, make quick sketches and use Pythagoras to find the
missing sides of each.
If cosx = 5/13, the cosine is + in I or IV
then sinx = 12/13 if x is in quad I
or sinx = -12/13 if x is in quad IV
if siny = 3/5, the sine is + ins I or II
then cosy = 4/5 if y is in quad I
or cosy = -4/5 if y is in quad II
You MUST know the expansions for cos(x+y) and sin(x-y)
I will do the first one:
cos(x+y) = cosxcosy - sinx siny
Case 1: x is in I, y is in I
cos(x+y) = cosxcosy - sinxsiny
= (5/13)(4/5) - (12/13)(3/5) = -16/65
case 2: x is in I, y is in II,
cos(x+y) = (5/13)(-4/5) - (12/13)(3/5)
= -56/65
[ check this: x= 67.4° , y = 143.1°
cos(67.4+143.1) = -.8615 or appr -56/65 ]
case 3: x is in IV , y is in I
cos(x+y) = (5/13)(4/5) - (-12/13)(3/5) = +56/65
case 4, x is in IV, y is in II
cos(x+y) = (5/13)(-4/5) - (-3/5)(12/13) = 16/65
Your conditions of x ≥ 0 was no problem, but
your second condition that y ≤ x/2 makes this more complicated
Look at my 4 cases:
case 1, x = 67.4° and y = 36.9
is 36.9 ≤ 33.7 ? NO, so case 1 has to be rejected
case 2, x is in I, y is in II
x = 67.4, y = 143.1 . Is y ≤ 1/2 x ??? NO, reject case 2
case 3, x is in IV , y is in I
x = 292.6, 36.9 , is y ≤ 1/2 x ? , YES, so case 3 works
case 4, x is in IV, y is in II
x = 292.6 , x = 143.1, is 143.1 ≤ (1/2)292.6 ?? YES, so case 4 works
so we have
case 3: x is in IV , y is in I
cos(x+y) = (5/13)(4/5) - (-12/13)(3/5) = +56/65
or
case 4, x is in IV, y is in II
cos(x+y) = (5/13)(-4/5) - (-3/5)(12/13) = 16/65
wheuwww! There has to be a better way, but that y ≤ x/2 is a weird restriction.
To determine the values of cos(x+y) and sin(x-y) using the given information, we can use the following trigonometric identities:
1) cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
2) sin(x - y) = sin(x)cos(y) - cos(x)sin(y)
a) To find cos(x+y), we need the values of cos(x) and sin(y).
Given:
cos(x) = 5/13
sin(y) = 3/5
We do not have the value of sin(x), but we can find it using the identity sin²(x) + cos²(x) = 1.
sin²(x) = 1 - cos²(x)
sin²(x) = 1 - (5/13)²
sin²(x) = 1 - 25/169
sin²(x) = 144/169
sin(x) = ±√(144/169)
sin(x) = ±12/13
Since 0 <= x and y <= x/2, sin(x) and sin(y) will be positive.
sin(x) = 12/13 (positive value)
Now, we can substitute the given values into the identity for cos(x + y):
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
cos(x + y) = (5/13)(cos(y)) - (12/13)(3/5)
cos(x + y) = (5/13)(cos(y)) - (36/65)
b) To find sin(x-y), we will again use the given values:
sin(x) = 12/13
cos(y) = 4/5
Substituting these values into the identity sin(x - y):
sin(x - y) = sin(x)cos(y) - cos(x)sin(y)
sin(x - y) = (12/13)(4/5) - (5/13)(3/5)
sin(x - y) = (48/65) - (15/65)
sin(x - y) = 33/65
Therefore, the final answers are:
a) cos(x + y) = (5/13)(cos(y)) - (36/65)
b) sin(x - y) = 33/65
To determine cos(x+y) and sin(x-y) using the given values of cos(x) and sin(y), we can use the trigonometric identities. Let's break down the steps:
a) To find cos(x+y), we can use the identity: cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)
Given that cos(x) = 5/13 and sin(y) = 3/5, we need to find cos(y) and sin(x).
To find sin(x), we can use the identity: sin^2(x) + cos^2(x) = 1.
Since cos(x) = 5/13, we can substitute it into the equation and solve for sin(x):
sin^2(x) + (5/13)^2 = 1
sin^2(x) + 25/169 = 1
sin^2(x) = 1 - 25/169
sin^2(x) = 144/169
sin(x) = ±√(144/169)
Since x is between 0 and π (or 0 and 180°), sin(x) should be positive. Therefore, sin(x) = √(144/169) = 12/13.
Now, we can substitute the values into the identity: cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y):
cos(x+y) = (5/13)*cos(y) - (12/13)*(3/5)
cos(x+y) = (5/13)*cos(y) - 36/65
To determine cos(y), we can use the identity: sin^2(y) + cos^2(y) = 1.
Since sin(y) = 3/5, we can substitute it into the equation and solve for cos(y):
(3/5)^2 + cos^2(y) = 1
9/25 + cos^2(y) = 1
cos^2(y) = 1 - 9/25
cos^2(y) = 16/25
cos(y) = ±√(16/25)
Since y is between 0 and π/2 (or 0 and 90°), cos(y) should be positive. Therefore, cos(y) = √(16/25) = 4/5.
Finally, we substitute cos(y) = 4/5 into the equation for cos(x+y):
cos(x+y) = (5/13)*(4/5) - 36/65
cos(x+y) = 20/65 - 36/65
cos(x+y) = -16/65
Therefore, cos(x+y) = -16/65.
b) To find sin(x-y), we can use the identity: sin(x-y) = sin(x)*cos(y) - cos(x)*sin(y)
Using the given values, we substitute sin(x) = 12/13, cos(y) = 4/5, cos(x) = 5/13, and sin(y) = 3/5 into the equation:
sin(x-y) = (12/13)*(4/5) - (5/13)*(3/5)
sin(x-y) = 48/65 - 15/65
sin(x-y) = 33/65
Therefore, sin(x-y) = 33/65.