Question

please help me solve!

tan(cos^-1(3/5))

Answer 1

start by drawing yourself a right triangle and let y = arccos(3/5)

Answer 2

I think it should be useful to use the fact,
\[1+\tan^2x=\frac{1}{\cos^2x}\]

Answer 3

where does the arc come from? just because there are two trig functions??

Answer 4

arccos(x) is another way of writing cos^-1(x)

Answer 5

but how would i do the math to solve? would it be arccos(3/5)?

Answer 6

but then what about tangent?

Answer 7

When you use arccos(3/5) you are literally saying,
"Then angle whose cosine is 3/5"
So cos x=3/5, but you don't know x. You don't need to know it. Now apply the formula I said before,
\[1+\tan^2x=\frac{1}{\cos^2x}\]
As you want the value of tan x, you need to simplify,
\[\tan x=\sqrt{\frac{1}{\cos^2x}-1}\] Or
\[\tan x=\sqrt{\frac{1}{(3/5)^2}-1}\]
And that is the solution.

Answer 8

i got 1.217054723035±πn,4.358644723035±πn.... i don't feel like that is right though

Answer 9

You should obtain, for tan x,
\[tan x=\sqrt{\frac{25}{9}-1}=\sqrt{\frac{16}{9}}=\pm\frac{4}{3}\]
You don't need the value x.

Answer 10

so you literally just did 3/5 * 3/5 and then subtract 1?

Answer 11

would that be the same for all of the functions?? for example the next problem is: sin(tan^-1(1/2)). would i just do sqrt(1/(1/2)^2 -1?

Answer 12

Exactly, because cos x=3/5. So (cos x)^2=(3/5)^2=9/25. In the case of the sine is a little different.

Answer 13

You should use the relation.
\[1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}\] But in this case you need the sine,
\[\sin x=\sqrt{\frac{1}{1+1/\tan^2x}}=\sqrt{\frac{1}{1+1/(1/2)^2}}\]

Answer 14

The result should be,
\[\sin x=\pm\frac{1}{\sqrt{5}}\]

Answer 15

oh i see, that is starting to make sense! sorry to continue to bother you but can you explain how i would do cosine? as an example: cos(sin^-1(2/3))

Answer 16

Yes, of course. The case with sine and cosine is easier. As you can use the relation,
\[\cos^2x+\sin^2x=1\] And solve for the cosine in the case you propose,
\[\cos x=\sqrt{1-\sin^2x}=\sqrt{1-(2/3)^2}=\pm \sqrt{5}/3\]

Answer 17

oh wow that one has less fractions in it! thank you so much for all of your help!

Answer 18

you need to decide whether it's positive or negative

Answer 19

how do i do that?

Answer 20

you need to know which quadrant the angle is in

Answer 21

You need to know by your problem, in which quadrant is your angle. If the problem don't give you any more data, then you must use the two solutions.

Answer 22

tan(cos^-1(3/5)) has an angle in the first quadrant. It's not explicitly stated but it's implied.
So the answer to the original question is +4/3

Answer 23

I'm sorry but I'm confused now, how do i know what the angle is to determine whether it is negative or positive

Answer 24

what is the range of angles that cos^-1 gives?

Answer 25

i really don't know! :O

Answer 26

You need more data, I think. In the case of the cos x= 3/5, the fraction 3/5 is the same as -3/(-5), but they don't represent the same quadrant.

Now suppose the problem ask you for,
\[\sin(\cos^{-1}(3/5))\] And adds that the angle is in the fourth quadrant. Then, the solution you must choose is with - sign, as you must know that sine is - and cosine is + for the fourth quadrant so the tangent is negative.

Now suppose the problem ask you for,
\[\sin(\cos^{-1}(3/5))\] And adds that the angle is in the first quadrant. Then, the solution you must choose is with + sign, as you must know that sine and cosine of the first quadrant are both positive, so the tangent is positive.

Answer 27

it's between 0 and 180 degrees. But since arccos(3/5) is positive, cosine is also positive, it's in first quadrant. It can not be the second quadrant because cosine is negative there.
And so tangent is positive in the first quadrant

Answer 28

Here, you have a brief graphic explanation,
http://www.mathmistakes.info/facts/TrigFacts/learn/uc/sign.html

Answer 29

cos(theta) = -3/-5 is in 3rd quadrant, which is impossible in this case

Answer 30

Yes, it's true. But the examples are okay. ;)

Answer 31

thank you both so much!! :)

Answer 32

Ok, I hope we haven't messed you so much. ;)

Answer 33

lol its okay I'm just trying to make sense of everything now :p