If a and B are two angles in Quadrant 2 such that tan a=-1/2 and tan B= -2/3, find cos(a+b)?
tan(a) = -1/2
oppsite side = 1: adjacentside = 2
hypotenuse = sqrt(1+4) = sqrt(5)
sin(a) = 1/�ã5
cos(a) = -2/�ã5
tan(b) = -2/3
opposite side = 2 and adjacentside = 3
hypotenuse = sqrt(4+9) = �ã13
sin(b) = 2/�ã13
cos(b) = -3/�ã13
cos(a+b) = cosa cosb - sina sinb
=(-2/�ã5)(-3/�ã13) - (1/�ã5)(2/�ã13)
= 6/�ã65 - 2/�ã65
= 4/�ã65
right?
I didn't check all, but you have the technique. You can easily check these on your calculator.
Yes, your calculations are correct. To find cos(a+b), you first need to find the values of sin(a), cos(a), sin(b), and cos(b) using the given information.
For angle a:
- tan(a) = opposite side/adjacent side = -1/2
- opposite side = 1, adjacent side = 2
- hypotenuse = sqrt(1+4) = sqrt(5)
- sin(a) = opposite side/hypotenuse = 1/�ã5
- cos(a) = adjacent side/hypotenuse = -2/�ã5
For angle b:
- tan(b) = opposite side/adjacent side = -2/3
- opposite side = 2, adjacent side = 3
- hypotenuse = sqrt(4+9) = �ã13
- sin(b) = opposite side/hypotenuse = 2/�ã13
- cos(b) = adjacent side/hypotenuse = -3/�ã13
Now, you can use the formula cos(a+b) = cosa cosb - sina sinb to find the value of cos(a+b). Substitute the values you found above:
cos(a+b) = (-2/�ã5)(-3/�ã13) - (1/�ã5)(2/�ã13)
= 6/�ã65 - 2/�ã65
= 4/�ã65
Therefore, cos(a+b) = 4/�ã65.