Inverse Trig Functions

**Chicken Enchilada** · September 11th 2009, 08:46 PM

I'm not getting this stuff at all. Finding math help online is quite hard, so I hope you guys can help me.

How do you solve problems like these ones:

$sin(arccos(3/5) + arccos(5/13))$

and

$cos(2arcsin(-2/3))$

How do you find the arccos of something like (3/5) without a calculator?!

I have some more questions, but I'll post them later for sake of clarity.

Thanks Everyone

**Grandad** · September 12th 2009, 12:26 AM

Hello Chicken Enchilada

Welcome to Math Help Forum.

You've come to the right place!

Originally Posted by Chicken Enchilada

I'm not getting this stuff at all. Finding math help online is quite hard, so I hope you guys can help me.

How do you solve problems like these ones:

$sin(arccos(3/5) + arccos(5/13))$

and

$cos(2arcsin(-2/3))$

How do you find the arccos of something like (3/5) without a calculator?!

I have some more questions, but I'll post them later for sake of clarity.

Thanks Everyone

To solve problems involving inverse trig functions, like $\arccos(\tfrac35)$ , let $x$ (or some variable) equal the inverse function and then express it the other way round; that is, in terms of the original trig function. So here:

Let $x = \arccos(\tfrac35)$ and $y = \arccos(\tfrac{5}{13})$ .

Then $\cos x = \tfrac35$ and $\cos y = \tfrac{5}{13}$ , and we want to find $\sin(x+y)$

Now expand: $\sin(x+y) = \sin x\cos y+ \cos x \sin y$

We know $\cos x$ and $\cos y$ . So all that remains is to find $\sin x$ and $\sin y$ , which we can do using Pythagoras' Theorem (or $\cos^2x + \sin^2x = 1$ , which is essentially the same thing).

So, $\sin x = \tfrac45$ (Can you see why?)

and $\cos x = ...$ ?

So $\sin(x+y) = ...$ ?

Can you complete it now?

(And for the second one, use $\cos2x = 1 -2\sin^2x$ )

Grandad

**Chicken Enchilada** · September 12th 2009, 08:39 AM

Originally Posted by Grandad

So, $\sin x = \tfrac45$ (Can you see why?)

Originally Posted by Grandad

$\cos x = ...$ ?

I assume you mean $sin(y)$ ? $cos(x) = (3/5)$ .

I got $siny = (12/13)$ . Plugged that in.....

$(4/5)(5/13) + (12/13)(3/5) = (56/65)$

Viola!

and for the second problem:

$cos(2x)$ Abstract the $arcsin(-2/3)$ .
$x = arcsin(-2/3)$

$sin(x) = (-2/3)$ Put it in terms of $sin(X)$ .

Plug it in and solve.
$cos(2x) = 1 - 2(-2/3)^2$

$(9/9) - (8/9) = (1 / 9)$

Yes!

This makes sense now. My Prof is one of those old man Profs that just goes up to the board and starts writing math equations. I can kind of see where he is coming from when he gave us stuff like $sin(arcos(x)) = \sqrt (1-x^2)$ . It looks similar to the

identity.

What are the least amount of theorems/identities that I need to know? I can permute them as needed (I think). Here are two so far.

The other questions I have concern derivatives and integrals so I'll post that in the Calc section.

Thanks a bunch Grandad!

**Soroban** · September 12th 2009, 12:19 PM

Hello, Chicken Enchilada1

What are the least amount of theorems/identities that I need to know?
I can permute them as needed (I think). . . . . Good for you!

Here are two so far:

. . $[1]\;\;\sin^2\!\theta + \cos^2\!\theta \;=\;1$

. . $[2]\;\;\cos2\theta \;=\;1-2\cos^2\!\theta$

The first one, [1], is very important.

. . Its two variations are good to know: . $\begin{array}{c}\tan^2\!\theta + 1 \;=\;\sec^2\!\theta \\ \cot^2\!\theta + 1 \;=\;\csc^2\!\theta \end{array}$

The second one, [2], has three variations: . $\cos2\theta \;=\;\begin{Bmatrix}2\cos^2\!\theta - 1 \\ 1 - 2\sin^2\!\theta \\ \cos^2\!\theta - \sin^2\!\theta \end{Bmatrix}$

If you know one of them, you can derive the other two (by substituting [1]),
. . but knowing all three saves a lot of time.

There is a third one you should know: . $[3]\;\;\sin2\theta \;=\;2\sin\theta\cos\theta$

Then there are these identities from basic trig:

. . $\begin{array}{c} \csc\theta \:=\:\dfrac{1}{\sin\theta} \\ \\[-3mm] \sec\theta \:=\:\dfrac{1}{\cos\theta} \\ \\[-3mm] \cot\theta \:=\:\dfrac{1}{\tan\theta} \end{array}\qquad\quad \begin{array}{c}\tan\theta \:=\:\dfrac{\sin\theta}{\cos\theta} \\ \\[-3mm] \cot\theta \:=\:\dfrac{\cos\theta}{\sin\theta} \end{array}$

**Grandad** · September 12th 2009, 12:32 PM

Hello Chicken Enchilada

Originally Posted by Chicken Enchilada

I assume you mean $sin(y)$ ? $cos(x) = (3/5)$ .

Yes of course I did. Sorry - but you've got it sorted in spite of me!

What are the least amount of theorems/identities that I need to know? I can permute them as needed (I think). Here are two so far.

The other questions I have concern derivatives and integrals so I'll post that in the Calc section.

Thanks a bunch Grandad!

That's a big question. I can't count the number of trig identities that I've used/learned over the years, but I'm afraid it probably runs to a few dozen. I couldn't say what the least number you'll need is - it depends very much on the level that you need to go to.

Just to give you some idea, here's a Wikipedia page: List of trigonometric identities - Wikipedia, the free encyclopedia. But don't let this put you off - it's like learning any topic: just do what you need to at any given time. And if you're going to use this stuff regularly, you'll just pick it up bit by bit. As an art teacher I once knew said: practice, patience and perseverance!

Grandad

**HallsofIvy** · September 12th 2009, 07:08 PM

Great Chicken Enchilada, but who is Viola ?

**Chicken Enchilada** · September 12th 2009, 08:27 PM

Thanks guys.

Originally Posted by HallsofIvy

Great Chicken Enchilada, but who is Viola ?

La voici!

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