Math Is Fun Forum

It might probably be helpful to draw the reference triangles, and then thinking of arccos and arcsin as "the angle with the given sin or cos value", which you can read off of the reference triangle.

It's also helpful to notice that:

which is the value that is easiest to see in the 45 degree angle reference triangle.

Thanks
Although it's arguable which of these forms is "simplest"!
For taking a derivative or integral, the final form is certainly most conveniant.

I guess you can also say:

2sqrt(2) - sqrt(2y) -2y +ysqrt(y) = 2^(3/2) - (2y)^(1/2) - 2y + y^(3/2)

The end results all are correct as far as I can tell.

I didn't check the intermediate steps of the second part, as it seems to work out simply as:

d/dx sec^n(x) = n sec ^(n-1) * sec(x) tanx  = n sec^n(x) tanx

That's what I got as well.

You can always check these like this, by making sure this equation holds:

quotient + (remainded/divisor) = (dividend/divisor)

At the very start all I did was add 1 and 4/u to get (u + 4)/u. The steps are all simple if you do them one at a time.

Thanks! That's great. Let's see:

Could you post an example of creating piecewise functions? None of the various ways I've found online for creating these seems to work here!

Hi Payari and welcome !

You need to use the polynomial long division process... it's hard to type that here, probably you have the procedure explained in a textbook..
If not there are sites online that explain it, for example:
http://www.sosmath.com/algebra/factor/fac01/fac01.html

For the cubics, you need to use the factor theorem, which states:

let P(x) be a cubic

(x - a) is a factor of P(x)   if and only if  P(a) = 0.

so for your example:

You need to try to test some values that will make this equal to 0. Let's try 1, -1, 0, 2, -2 or such.
Usually it is easy since the questions in books are designed to be
I usually try -2 first and look:

-8 + 8 -8 + 8 = 0 success!

Therefore, P(-2) = 0. By the factor theorem:

x -- 2 = (x + 2) is a factor of P(x)

Now you use polynomial long division, divide x + 2 into P(x).

Your result will be a quadratic factor, lets call it Q(x).

Then P(x) = (x + 2)Q(x)

and you can factor Q(x) using the standard methods.

For the integral:

Notice that

And here it just so happens that cos(x) is the derivative of sin(x).

So we get:

And now the definite integral is easy to evaluate by the fundamental theorem of calculus.