SOLUTION: Suppose sinA = 12/13 with 90�&#8804;A&#8804;180�. Suppose also that sinB = -7/25 with -90�&#8804;B&#8804;0�. Find cos(A

Click here to see ALL problems on Trigonometry-basics

Question 619835: Suppose sinA = 12/13 with 90�≤A≤180�. Suppose also that sinB = -7/25 with -90�≤B≤0�. Find cos(A - B).
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
Suppose sinA = 12/13 with 90�≤A≤180�. Suppose also that sinB = -7/25 with -90�≤B≤0�. Find cos(A - B).
**
cos addition formula: Cos(A-B)=cosAcosB+sinAsinB
given sin A=12/13=opposite side/hypotenuse
adjacent side=√(13^2-12^2)=√(169-144)=√25=5
cos A=adjacent side/hypotenuse=-5/13 (in quadrant II where cos<0)
..
given sin B=-7/25=opposite side/hypotenuse
adjacent side=√(25^2-7^2)=√(625-49)=√576=24
cos B=adjacent side/hypotenuse=24/25 (in quadrant IV where cos>0)
..
Cos(A-B)=-5/13*24/25+12/13*+7/25=-120/325+84/325=-36/325≈-.1108

SOLUTION: Suppose sinA = 12/13 with 90�&#8804;A&#8804;180�. Suppose also that sinB = -7/25 with -90�&#8804;B&#8804;0�. Find cos(A - B).

SOLUTION: Suppose sinA = 12/13 with 90�≤A≤180�. Suppose also that sinB = -7/25 with -90�≤B≤0�. Find cos(A - B).