Finding Value of Inverse Trig Funct

In summary, the person is trying to solve for cos(arctan(2)+arctan(3)) and is not sure what numbers to use. They think that arctan(2) is equivalent to tanx=2 and arctan(3) is equal to tanx=3. They use the identity cos(A+B) to solve for x and y.
  • #1
vipertongn
98
0

Homework Statement



find the exact value of cos(arctan(2)+arctan(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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  • #2
vipertongn said:
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.
vipertongn said:
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.
 
  • #3
what do you mean?
 
  • #4
vipertongn said:
what do you mean?

Try using the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
 
  • #5
How would i solve for x and y then? because like thinking it in terms of sin/cos=2 or 3 seems kinda hard
 
  • #6
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?
 
  • #7
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:
  • #8
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?
 
  • #9
ArcTan(3) + ArcTan(2) = 3Pi/4
 
  • #10
vipertongn said:
do i use sine(2a)=2sinacosa?

That would be a straightforward method. :smile:
 
  • #11
Thanks so much for all your help ^^.
 

1. What are inverse trigonometric functions?

Inverse trigonometric functions are mathematical functions that are used to find the angle measure of a right triangle when given the ratio of two sides. They are the inverse operations of the basic trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant).

2. Why do we need to find the value of inverse trigonometric functions?

Inverse trigonometric functions are useful in solving real-world problems involving angles and sides of triangles. They are also commonly used in calculus and other advanced math topics.

3. How do we find the value of inverse trigonometric functions?

The value of an inverse trigonometric function can be found using a calculator or by using trigonometric identities and equations. It is important to follow the correct order of operations and use the appropriate inverse function for the given problem.

4. Can inverse trigonometric functions have multiple solutions?

Yes, inverse trigonometric functions can have multiple solutions. This is because a single ratio can correspond to multiple angles in a right triangle. It is important to know which quadrant the angle is in and to use the correct sign for the solution.

5. What are the domain and range of inverse trigonometric functions?

The domain of inverse trigonometric functions is the set of possible input values (ratios) and the range is the set of possible output values (angles). Depending on the function, the domain and range can vary, but they are usually limited to certain intervals to ensure a one-to-one relationship between the input and output values.

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