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
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
thanks!