Steve
answered
if A is in QII and sinA = 12/13, the cosA = -5/13
if B is in QI and cosB = 15/17, then sinB = 8/17
So,
sin(A-B) = sinAcosB - cosAsinB
= (12/13)(15/17) - (-5/13)(8/17)
= 180/221 + 40/221 = 220/221
Explain Bot
answered
sin(a - B) = sin(a)cos(B) - cos(a)sin(B)
Given the information you provided:
- sin(a) = 12/13 (a lies in quadrant II, where sin(a) is positive)
- cos(B) = 15/17 (B lies in quadrant I, where cos(B) is positive)
To find cos(a), we can make use of the Pythagorean Identity:
cos^2(a) + sin^2(a) = 1
Since sin(a) is given as 12/13 and a lies in quadrant II, we can deduce that cos(a) must be negative. Solving for cos(a):
cos^2(a) = 1 - sin^2(a)
cos^2(a) = 1 - (12/13)^2
cos^2(a) = 1 - 144/169
cos^2(a) = 25/169
cos(a) = -5/13 (taking the negative value as cos(a) is negative in quadrant II)
Similarly, to find sin(B), we can use the same approach with the Pythagorean Identity:
cos^2(B) + sin^2(B) = 1
Since cos(B) is given as 15/17 and B lies in quadrant I, we can deduce that sin(B) must be positive. Solving for sin(B):
sin^2(B) = 1 - cos^2(B)
sin^2(B) = 1 - (15/17)^2
sin^2(B) = 1 - 225/289
sin^2(B) = 64/289
sin(B) = 8/17 (taking the positive value as sin(B) is positive in quadrant I)
Now that we have the values for sin(a), cos(B), sin(B), and cos(a), we can substitute them into the trigonometric identity:
sin(a - B) = sin(a)cos(B) - cos(a)sin(B)
sin(a - B) = (12/13)(15/17) - (-5/13)(8/17)
sin(a - B) = (180/221) + (40/221)
sin(a - B) = 220/221
Therefore, sin(a - B) = 220/221.
Step-by-Step Bot
answered
Step 1: First, find sin(B) using the given information:
Since cos(B) = 15/17 and B lies in quadrant I, we know that sin(B) is positive. To find sin(B), we can use the Pythagorean identity: sin^2(B) + cos^2(B) = 1.
Plugging in the value of cos(B), we have:
sin^2(B) + (15/17)^2 = 1
sin^2(B) + 225/289 = 1
sin^2(B) = 1 - 225/289
sin^2(B) = (289 - 225)/289
sin^2(B) = 64/289
Taking the square root of both sides, we get:
sin(B) = ±8/17
Since B is in quadrant I and sin(B) is positive, we have sin(B) = 8/17.
Step 2: Next, find cos(a) using the given information:
We know that sin(a) = 12/13, and a is in quadrant II. In quadrant II, cos(a) is negative. We can use the Pythagorean identity to find cos(a):
sin^2(a) + cos^2(a) = 1
(12/13)^2 + cos^2(a) = 1
144/169 + cos^2(a) = 1
cos^2(a) = 1 - 144/169
cos^2(a) = (169 - 144)/169
cos^2(a) = 25/169
Taking the square root of both sides and considering that cos(a) is negative in quadrant II, we have:
cos(a) = -5/13
Step 3: Finally, use the trigonometric addition formula to find sin(a - B):
The trigonometric addition formula for sin is: sin(a - B) = sin(a) * cos(B) - cos(a) * sin(B)
Plugging in the values we found earlier, we have:
sin(a - B) = (12/13) * (15/17) - (-5/13) * (8/17)
sin(a - B) = (180/221) + (40/221)
sin(a - B) = 220/221
Therefore, sin(a - B) = 220/221.
