If A is between 270degrees and 360degrees, and cos A=12/13, find the exact value of:
sin2a, cos2a, sin1/2a, cos1/2a
since acos(12/13) = .39479, a = 2*pi - .39479 = 5.88839
2a = 11.77678 is in the 4th quadrant, so
sin 2a = 2sin(a)cos(a) = 2(-5/13)(12/13) = -120/169
cos(2a) = cos^2(a) - sin^2(a) = 144/169 - 25/169 = 119/169
a/2 is in the 2nd quadrant
sin(a/2) = √(1/26)
cos(a/2) = -√(25/26)
if cosA = 12/13 in quadr IV, then
sinA = -5/13
sin 2A = 2sinAcosA = 2(12/13)(-5/13) = -120/169 =
cos 2A = cos^2 A - sin^2 A
= 144/169 - 25/169 = 119/169
we know cos2A = 1 - 2sin^2 A
or
cos A = 1 - 2sin^2 (A/2)
12/13 = 1 - 2sin^2 (A/2)
2sin^2 (A/2) = 1 - 12/13 = 1/13
sin^2 (A/2) = 1/26
sin (A/2) = ±1/√26
but if A is in IV, then A/2 is in II
so sin(A/2) = 1/√26
using
cos A = 2cos^2 (A/2) -1
try the last one.
To find the values of sin(2A), cos(2A), sin(A / 2), and cos(A / 2), we will use trigonometric identities and the given information.
Step 1: Determine the value of sin(A). Since A is between 270 degrees and 360 degrees and cos(A) = 12/13, we can use the Pythagorean identity:
sin^2(A) + cos^2(A) = 1
sin^2(A) + (12/13)^2 = 1
sin^2(A) = 1 - (12/13)^2
sin^2(A) = 1 - 144/169
sin^2(A) = 25/169
Taking the square root of both sides:
sin(A) = ± √(25/169)
sin(A) = ± 5/13
Since A is between 270 degrees and 360 degrees, sin(A) cannot be negative, so sin(A) = 5/13.
Step 2: Use the double-angle formulas to find sin(2A) and cos(2A).
sin(2A) = 2sin(A)cos(A)
sin(2A) = 2(5/13)(12/13)
sin(2A) = 120/169
cos(2A) = cos^2(A) - sin^2(A)
cos(2A) = (12/13)^2 - (5/13)^2
cos(2A) = 144/169 - 25/169
cos(2A) = 119/169
Therefore, sin(2A) = 120/169 and cos(2A) = 119/169.
Step 3: Use the half-angle formulas to find sin(A/2) and cos(A/2).
sin(A/2) = ± √((1 - cos(A))/2)
sin(A/2) = ± √((1 - 12/13)/2)
sin(A/2) = ± √(1/26)
sin(A/2) = ± √(1) / √(26)
sin(A/2) = ± 1 / √(26)
Since A is between 270 degrees and 360 degrees, sin(A/2) cannot be negative, so sin(A/2) = 1 / √(26).
cos(A/2) = ± √((1 + cos(A))/2)
cos(A/2) = ± √((1 + 12/13)/2)
cos(A/2) = ± √(25/26)
cos(A/2) = ± √(25) / √(26)
cos(A/2) = ± 5 / √(26)
Since A is between 270 degrees and 360 degrees, cos(A/2) cannot be negative, so cos(A/2) = 5 / √(26).
Therefore, sin(A/2) = 1 / √(26) and cos(A/2) = 5 / √(26).
To find the exact values of sin(2A), cos(2A), sin(A/2), and cos(A/2), we first need to determine the value of sin(A) and cos(A).
Given that cos(A) = 12/13, we can use the Pythagorean identity to find sin(A):
sin(A) = √(1 - cos^2(A))
= √(1 - (12/13)^2)
= √(1 - 144/169)
= √(169/169 - 144/169)
= √(25/169)
= 5/13
Now that we have the values of sin(A) and cos(A), we can proceed to find the desired values.
1. sin(2A):
We can use the double angle identity for sine:
sin(2A) = 2sin(A)cos(A)
= 2 * (5/13) * (12/13)
= 120/169
2. cos(2A):
We can use the double angle identity for cosine:
cos(2A) = cos^2(A) - sin^2(A)
= (12/13)^2 - (5/13)^2
= 144/169 - 25/169
= 119/169
3. sin(A/2):
We can use the half angle formula for sine:
sin(A/2) = ±√((1 - cos(A))/2)
= ±√((1 - 12/13)/2)
= ±√(1/13)
= ±(1/√13)
= ±(√13/13)
Since A is between 270° and 360°, A/2 will be between 135° and 180°. Thus, sin(A/2) must be positive, so the answer is √13/13.
4. cos(A/2):
We can use the half angle formula for cosine:
cos(A/2) = ±√((1 + cos(A))/2)
= ±√((1 + 12/13)/2)
= ±√(25/26)
= ±(5/√26)
= ±(5√26/26)
Since A is between 270° and 360°, A/2 will be between 135° and 180°. Thus, cos(A/2) must be negative, so the answer is -5√26/26.
Therefore, the exact values are:
sin(2A) = 120/169
cos(2A) = 119/169
sin(A/2) = √13/13
cos(A/2) = -5√26/26