If cos(a)=1/2 and sin(b)=2/3, find sin(a+b), if
1) Both angles are acute; Answer: (sqrt(15)+2)/6
ii) a is an acute angle and pi/2 < b < pi; Answer: (2-sqrt(15))/6
2. Find the exact value of the six trigonometric functions of 13pi/12.
Partial answer: cos(13pi/12)=-(sqrt(6)+sqrt(2))/4
13/12 = 5/4 - 1/6
sin 5pi/4 = -1/√2
cos 5pi/4 = -1/√2
sin pi/6 = 1/2
cos pi/6 = √3/2
so,
sin(13pi/12) = sin(5pi/4 - pi/6) = (-1/√2)(√3/2)-(-1/√2)(1/2) = (1-√3)/√8
and so on for the others
To find the value of sin(a+b) in the given scenarios, we can use the trigonometric identity: sin(a+b) = sin(a)cos(b) + cos(a)sin(b).
1) Both angles are acute:
Given that cos(a) = 1/2 and sin(b) = 2/3, we need to find sin(a) and cos(b).
To find sin(a), we can use the Pythagorean identity: sin^2(a) + cos^2(a) = 1. Using the value of cos(a) = 1/2, we have:
sin^2(a) + (1/2)^2 = 1
sin^2(a) + 1/4 = 1
sin^2(a) = 1 - 1/4
sin^2(a) = 3/4
sin(a) = sqrt(3)/2
To find cos(b), we can use the Pythagorean identity: sin^2(b) + cos^2(b) = 1. Using the value of sin(b) = 2/3, we have:
(2/3)^2 + cos^2(b) = 1
4/9 + cos^2(b) = 1
cos^2(b) = 1 - 4/9
cos^2(b) = 5/9
cos(b) = sqrt(5)/3
Now, substituting these values in the expression sin(a+b) = sin(a)cos(b) + cos(a)sin(b):
sin(a+b) = (sqrt(3)/2)(sqrt(5)/3) + (1/2)(2/3)
sin(a+b) = sqrt(15)/6 + 1/3
sin(a+b) = (sqrt(15)+2)/6
ii) a is an acute angle and pi/2 < b < pi:
Given that cos(a) = 1/2 and sin(b) = 2/3, we need to find sin(a) and cos(b).
To find sin(a), we can use the Pythagorean identity: sin^2(a) + cos^2(a) = 1. Using the value of cos(a) = 1/2, we have:
sin^2(a) + (1/2)^2 = 1
sin^2(a) + 1/4 = 1
sin^2(a) = 1 - 1/4
sin^2(a) = 3/4
sin(a) = sqrt(3)/2
To find cos(b), we can use the Pythagorean identity: sin^2(b) + cos^2(b) = 1. Using the value of sin(b) = 2/3, we have:
(2/3)^2 + cos^2(b) = 1
4/9 + cos^2(b) = 1
cos^2(b) = 1 - 4/9
cos^2(b) = 5/9
cos(b) = sqrt(5)/3
Now, substituting these values in the expression sin(a+b) = sin(a)cos(b) + cos(a)sin(b):
sin(a+b) = (sqrt(3)/2)(sqrt(5)/3) + (1/2)(2/3)
sin(a+b) = (sqrt(15)/6) + 1/3
sin(a+b) = (2-sqrt(15))/6
2) To find the exact value of the six trigonometric functions of 13pi/12, we can use the unit circle and evaluate the trigonometric ratios at that angle.
Given that cos(13pi/12) = -(sqrt(6)+sqrt(2))/4, we can find the values of sin(13pi/12), tan(13pi/12), cot(13pi/12), sec(13pi/12), and csc(13pi/12) using the following trigonometric identities:
sin(theta) = sqrt(1 - cos^2(theta))
tan(theta) = sin(theta) / cos(theta)
cot(theta) = 1 / tan(theta)
sec(theta) = 1 / cos(theta)
csc(theta) = 1 / sin(theta)
We have already been given the value of cos(13pi/12). We can substitute it into the identity sin(theta) = sqrt(1 - cos^2(theta)) to find sin(13pi/12).
sin(13pi/12) = sqrt(1 - (-(sqrt(6)+sqrt(2))/4)^2)
sin(13pi/12) = sqrt(1 - (6 + 2 + 2sqrt(12))/16)
sin(13pi/12) = sqrt(1 - (4 + sqrt(12))/8)
sin(13pi/12) = sqrt(1 - (2 + sqrt(3))/4)
sin(13pi/12) = sqrt((4 - 2 - sqrt(3))/4)
sin(13pi/12) = sqrt((2 - sqrt(3))/4)
sin(13pi/12) = sqrt(2 - sqrt(3))/2
Similarly, we can find the values of tan(13pi/12), cot(13pi/12), sec(13pi/12), and csc(13pi/12) using the trigonometric identities mentioned above.