1. ## Trigonometry

Given that sinA = 3/5 and cosB = 12/13, where A is obtuse and B is acute, find the exact values of cos(A+B) and cot(A-B)

I know that cos(A+B) = cosAcosB - sinAsinB and that I need to draw a triangles of some sort with the sides 3 and 5 on one and 12 and 13 on the other, but I don't know how to find the last length because I don't think they would be right-angled triangles.

I know that cot = 1/tan

Where should I go from here?

Thanks,
Delta

2. $\displaystyle 33 \over 65$ for the cos(a+b) because:

$\displaystyle A = arcsin(\frac{3}{5}) + 2 \pi$
$\displaystyle B = arccos(\frac{12}{13})$

and apply the relation you provided.

$\displaystyle 63 \over 16$ for the cot(a-b) because:

$\displaystyle A = arcsin(\frac{3}{5}) + 2 pi$
$\displaystyle B = arccos(\frac{12}{13})$

$\displaystyle cot(A - B) = -tan( arcsin(\frac{12}{13}) + arcsin(\frac{3}{5}) )$

and applying this:
$\displaystyle tan(A+B) = \frac{cos(A) sin(B) + sin(A) cos(B)}{cos(A) cos(B) - sin(A) sin(B)}$

3. or you can apply

$\displaystyle tan(A-B) = \frac{sin(A) cos(B) - cos(A) sin(B)}{cos(A) cos(B) + sin(A) sin(B)}$

4. You have the idea, except on the cot = 1/tan.

cot = 1/tan is correct, but that is not what you need here. You need here the trig identity cot(A-B), in the same idea that you showed the trig identity cos(A+B).

Any acute angle can be in a right triangle---it can be one of the two acute angles in a right triangle.
Any angle can be viewed as a reference angle, and a reference angle is an acute angle measured from the horizontal axis or from the x-axis.

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Angle A is an obtuse angle, so its more than 90 degrees but less than 180 degrees. Its terminal side then is in the 2nd quadrant. Its reference angle then is an acute angle based on the negative or lefthand side of the x-axis.

sinA = 3/5 = (opposite side)/hypotenuse = opp/hyp

Imagine the reference triangle of angle A. (A reference triangle is always a right triangle, whatever is the reference angle.)
Its reference angle is angle A.
Its opposite side (the leg of the right triangle that is opposite angle A) is 3.
Its hypotenuse is 5.

By Pythagorean theorem,
Angle A is in the 2nd quadrant, where x is negative.

Therefore,

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Angle B is acute, so it is in the 1st quadrant.

cosB = 12/13

Follow the procedure above, re angle A.

hyp = 13
so, opp = 5 ----in the 1st quadrant, all of the opp, adj and hyp are positive.

Then,
sinB = opp/hyp = 5/13

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Exact value of cos(A+B).

cos(A+B)
= cosAcosB -sinAsinB
= (-4/5)(12/13) -(3/5)(5/13)
= -48/65 -15/65

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Exact value of cot(A-B)

cot(A-B)
= [cotBcotA +1] / [cotB -cotA]
= [(12/5)(-4/3) +1] / [12/5 -(-4/3)]
= [-48/15 +1 ]/ [12/5 +4/3]
= [(-48 +15)/15] / [(36 +20)/15]
= (-33/15) / (56/15)
= (-33/15)*(15/56)