Find the exact value of:

cos(arctan(2) + arctan(3))

Please explain how you got your answer. I have the answer, but I don't know how to get to it.

2 answers

this resembles the cos(A+B) expansion
cos(A+B) = cosAcosB - sinAsinB

so cos(arctan(2) + arctan(3))
= cos(arctan2)cos(arctan3) - sin(arctan2)sin(arctan3)

let's look at arctan2,
draw right-angled triangle in the first quadrant, with a base (adjacent) of 1 and a height (opposite) of 2
Then the hypotenuse is √5
so the cos(angle) = 1/√5 and sin(angle) = 2/√5

sofar you have
(1/√5)cos(arctan3) - (2/√5)sin(arctan3)

then for arctan3, again draw a right-angled triangle but this time the height is 3, so the hypotenuse is √10

then the sine of that angle is 3√10 and cosine of that angle is 1/√10

so your expression of cos(arctan2)cos(arctan3) - sin(arctan2)sin(arctan3)
= (1/√5)(1/√10) - (2/√5)(3/√10)
= 1/√50 - 6/√50
= -5/√50
= -5/(5√2)
= -1/√2 or -√2/2 after rationalizing

BTW, we could check this:
arctan2 + arctan3 = 135º
and cos 135º = -.707.. = - √2/2
thanks!
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