Inverses of Trig Functions: Tan^-1sq. root of 3, etc.

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Violagirl

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Hi, I'm having trouble understanding to find the invereses of sine, cosine, and tangent. I was wondering if someone could show me how to do a few so I can get a better understanding on how to do them.
Find the exact value of each expression.

1. Tan^-1sq. root of 3.

2. Cos^-1(cos 4pi/5)

3. sin(sin^-1 1/4)
 
Re: Inverses of Trig Functionst

\(\displaystyle \tan^{-1}\sqrt{3}\) is the angle whose tangent is ?3 or ?3/1.

In quadrant I draw a triangle so that your central angle has its opposite side about ?3 units and its adjacent side 1 unit. Use the Pythagorean Theorem to determine the length of the hypotenuse. You should recognize this to be a 30-60 degree right triangle. Therefore Tan[sup:14324vgy]-1[/sup:14324vgy]?3 = 60°. Us similar techniques to do the other two.
 
Re: Inverses of Trig Functionst

I take it back after looking at the other two more closely. You don't use the same technique. The other two are similar to finding, say, the square of the square root of 2 or "what is the cube root of x cubed?".
 
Re: Inverses of Trig Functionst

Ah okay. I'm a little confused on that, could I see an example?
 
Re: Inverses of Trig Functionst

sec (sec[sup:108w3ts2]-1[/sup:108w3ts2] 3) = 3

tan[sup:108w3ts2]-1[/sup:108w3ts2](tan 45°) = tan[sup:108w3ts2]-1[/sup:108w3ts2] 1 = 45°.
 
Re: Inverses of Trig Functionst

(2) and (3) are solved through the theorem:

\(\displaystyle f^{-1}[f(x)] \, = \, f[f^{-1}(x)] \, = \, x\)

so if you are working with say sine function then

\(\displaystyle \sin^{-1}[\sin(\theta)] \, = \, \sin[\sin^{-1}(\theta)] \, = \, \theta\)

or

\(\displaystyle \sin^{-1}[\sin(\frac{\pi}{4})] \, = \, \sin[\sin^{-1}(\frac{\pi}{4})] \, = \, \frac{\pi}{4}\)
 
Violagirl said:
Hi, I'm having trouble understanding to find the invereses of sine, cosine, and tangent. I was wondering if someone could show me how to do a few so I can get a better understanding on how to do them.
Find the exact value of each expression.

1. Tan^-1sq. root of 3.

This one has already been commented on.

2. Cos^-1(cos 4pi/5)

4?/5 is in Quadrant II. cos^-1 x has a range of [0, ?], so the solution is 4?/5.

3. sin(sin^-1 1/4)
Click to expand...
sin^-1 (1/4) would give a first quadrant angle, so the sin of that angle would be 1/4.

You need to check the domain and range of the values given, to be sure that the function and its inverse "cancel out".
 
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