Finding Value of Inverse Trig Funct

1. The problem statement, all variables and given/known data

find the exact value of cos(arctan(2)+arctan(3))

3. The attempt at a solution

I know that arctan(2) is equivalent to tanx=2 and arctan(3) is equal to tanx=3

I was thinking perhaps sinx/cosx=2 and 3 but I am not sure what numbers to use. I'm kind of weak in my trig rules =( can someone explain to me how I would go about this problem? I know the answer is -sqrt2/2 but I want to know HOW to do it.
 
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I know that arctan(2) is equivalent to tanx=2 and arctan(3) is equal to tanx=3
I would say tan(y) = 3 since they are different angles.
I was thinking perhaps sinx/cosx=2 and 3 but I am not sure what numbers to use.
It may be difficult to evaluate the cosine of the two angles if there's a quotient involved. Try separating the two angles by using the identities for adding two angles. Then you will just need to evaluate terms like cos(arctan(2)), for which you can draw a right triangle with a height of 2 and a base of 1. The angle of magnitude arctan(2) would be the angle adjacent to the base. You can then read the cosine right off the triangle.
 
what do you mean?
 
How would i solve for x and y then? because like thinking it in terms of sin/cos=2 or 3 seems kinda hard
 

robphy

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As slider142 says, use the identity for cos(A+B), where tan A=2 and tan B=3.
The idea now is to express the identity for cos(A+B) in terms of tan A and tan B.
For example, can you write cos A in terms of tan A?
 
OHhhh yes think of it like a triangle, the angle would be arctan(2) the sine would be equal to 2 and the cos would be equal to 1 and the hypotenuse is equal to sqrt 5. then when you use CAH it would be 1/sqrt(5) for cos(arctan(2) right?Yes i got the answer =D thanks
 
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I got a new problem now ummm sine(2arcsec(13/5))

I went about using the similar method above but without the identity. but because of the 2 there I don't know how to get the answer...

do i use sine(2a)=2sinacosa?
 
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ArcTan(3) + ArcTan(2) = 3Pi/4
 
Thanks so much for all your help ^^.
 

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