# Math Help - any help would be great

1. ## any help would be great

ok, thanks for taking a look for starters,
here we go,
i have two functions y= sinx-cosx and y=2cos2x
ive selected an area that has been "trapped" by these two functions and must find the domain and range of the area selected, how do i go about doing this
i aslo need to find the points of intersection of the two functions in the area i chose , do i start by making both functions equal each other?, if so how do i solve that?
thanks again

2. Hello, bssm19!

I have two functions: $y\:=\:\sin x - \cos x$ and $y\:=\:2\cos2x$
I've selected an area that has been "trapped" by these two functions
and must find the domain and range of the area selected.
How do i go about doing this?

Not a clue . . .
Since the area is some constant, I don't see how it can have a doman and range.

I also need to find the points of intersection of the two functions in the area i chose.
Do i start by making both functions equal each other? . . . Yes!
If so, how do i solve that? . . . It's not easy

I'll get you started . . .

Equate the two functions: . $2\underbrace{\cos2x}\;=\;\sin x - \cos x$

Double-angle identity: . $2(\overbrace{\cos^2x - \sin^2x})\;=\;\sin x - \cos x$

We have: . $2(\cos^2x - \sin^2x) + \cos x - \sin x \;= \;0$

Factor: . $2(\cos x - \sin x)(\cos x + \sin x) + (\cos x - \sin x)\;=\;0$

Factor: . $(\cos x - \sin x)\,\left[2(\cos x + \sin x) + 1\right]\;=\;0$

And we have two equations to solve:

$(a)\;\;\cos x - \sin x\:=\:0\quad\Rightarrow\quad \sin x\:=\:\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}$ $\:=\:1\quad\Rightarrow\quad\tan x\:=\:1$
. . . Hence: . $\boxed{x\:=\:\frac{\pi}{4} + \pi n}$ *

$(b)\;\;2(\cos x + \sin x) + 1\:-\:0\quad\Rightarrow\quad \sin x + \cos x \:=\:-\frac{1}{2}$

. . . Square both sides: . $\sin^2x + 2\sin x\cos x + \cos^2x\:=\:\frac{1}{4}$

. . . Since $\sin^2x + \cos^2x\:=\:1$ and $2\sin x\cos x\:=\:\sin2x$
. . . . we have: . $\sin2x + 1 \:= \:\frac{1}{4}\quad\Rightarrow\quad\sin2x\:=\:\text {-}\frac{3}{4}$

. . . Hence: . $2x\:=\:\sin^{-1}\!\left(\text{-}\frac{3}{4}\right) + 2\pi n\quad\Rightarrow\quad \boxed{x\:=\:\frac{1}{2}\sin^{-1}\!\left(\text{-}\frac{3}{4}\right) + \pi n}$ *

[Note: Since we squared the equation, check for extraneous roots.]

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* Of course, the value of $n$ depends on what area you selected.