# Finding Value of Inverse Trig Funct

#### vipertongn

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.

#### slider142

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.

#### vipertongn

what do you mean?

#### slider142

what do you mean?
Try using the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

#### vipertongn

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

##### timelike observer
Science Advisor
Homework Helper
Gold Member
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?

#### vipertongn

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

Last edited:

#### vipertongn

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?

#### Gregg

ArcTan(3) + ArcTan(2) = 3Pi/4

#### slider142

do i use sine(2a)=2sinacosa?
That would be a straightforward method.

#### vipertongn

Thanks so much for all your help ^^.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving