Trig equations

**Alan306090** · Jan 17th 2010, 02:17 PM

Hello all! I'm studying for a math cumulative on tuesday, and I've come across these problems that I cant solve.

How do I solve this between 0 inclusive and 2 pi?

Cos^2x + 2sinxcosx - sin^2 x = 0

sin(x +1) = cos x

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central along of sector is 50, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

On another note: If I have more trig questions, do I make a new post or continue to post them here?

Edit: I realized I posted this under the wrong section, sorry! Please move this to the trig section

**TKHunny** · Jan 17th 2010, 05:56 PM

On the first, a little experimentation shows a nice solution.

$\cos^{2}(x) + 2\cos(x)\sin(x) - \sin^{2}(x) =$

$\cos^{2}(x) + 2\cos(x)\sin(x) + \sin^2(x) - 2\sin^{2}(x) =$

$2\cos(x)\sin(x) + 1 - 2\sin^{2}(x) =$

$\sin(2x) + \cos(2x) =$

$\sqrt{2}\sin(2x + \frac{\pi}{4}) = 0$

Now what?

It's important to note that the first step was unnecessary. I was thinking of another route when I did that. This should stand as excellent encouragement to give it a try and see where it leads. If you end up with a path you had not imagined at first, so be it.

**TKHunny** · Jan 17th 2010, 06:02 PM

On third thought, that first step is not so bad. It leads to where I was originally headed.

$(\cos(x) + \sin(x))^{2} - 2\sin^2(x) = 0$

You should be able to factor that (difference of squares) and (after soem more algebra and trigonometric substitutions) get the same results.

Just keep your mind open.

**Soroban** · Jan 17th 2010, 07:45 PM

Hello, Alan306090!

Here's the first one . . .

Solve in the interval $[0,2\pi]$

. . $\cos^2\!x + 2\sin x\cos x - \sin^2\!x \:=\: 0$

$\text{We have: }\;\underbrace{(\cos^2\!x - \sin^2\!x)}_{\text{This is }\cos2x} + \underbrace{(2\sin x\cos x)}_{\text{This is }\sin2x} \;=\;0$

Then we have: . $\cos2x +\sin2x \:=\:0 \qquad\Rightarrow\qquad \sin2x \:=\:-\cos2x \qquad\Rightarrow\qquad\frac{\sin2x}{\cos2x} \;=\;-1$

Hence: . $\tan2x \:=\:-1 \qquad\Rightarrow\qquad 2x \;=\;\frac{3\pi}{4},\;\frac{7\pi}{4},\;\frac{11\pi }{4},\;\frac{15\pi}{4}$

Therefore: . $x \;=\;\frac{3\pi}{8},\;\frac{7\pi}{8},\;\frac{11\pi }{8},\;\frac{15\pi}{8}$

**Soroban** · Jan 17th 2010, 08:04 PM

Hello again, Alan306090!

Solve in the interval $[0,2\pi]:\quad\sin(x +1) \:=\: \cos x$

We have: . $\sin x\cos1 + \cos x\sin1 \:=\:\cos x \qquad\Rightarrow\qquad \sin x\cos 1 \;=\;\cos x - \cos x\sin 1$

Factor: . $\sin x\cos 1 \;=\;\cos x(1 - \sin 1) \qquad\Rightarrow\qquad \frac{\sin x}{\cos x} \;=\;\frac{1-\sin 1}{\cos 1}$

Hence: . $\tan x \;=\;\frac{1-\sin 1}{\cos 1} \qquad\Rightarrow\qquad x \;=\;\tan^{-1}\left(\frac{1-\sin1}{\cos1}\right)$

Therefore: . $x \;\approx\;\begin{Bmatrix}0.2854\\ 3.4270\end{Bmatrix}\text{radians}$

**Alan306090** · Jan 18th 2010, 12:29 PM

Thank you Soroban!

If I have more questions do I post them here, or make a new topic?

**TKHunny** · Jan 18th 2010, 03:09 PM

That division by cos(2x) is not a good plan. This changes the Domain of the equation.

Kudos for using the "first" form of cos(2x). I almost never use that form and just didn't see it. Please note how various approaches will cause you to learn more.

Alan306090, please also note how personal exploration and doing your own work will cause you to learn even more.

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