Gr. 12 Trig question

[JamesMclean]

Solve each equation algebraically for x where : 0 (Less than or equal to) x (less than or equal to) 2pi

Sin(x) + Cos(x) = 1
Sin(x) + Cos(x) = Sin(x)^2 + Cos(x)^2
Sin(x)^2 + Cos(x)^2 - Sin(x) - Cos(x) = 0

Then I don't know what to do. Looks easy, feels like I'm just missing something...

 

[tkhunny]

Neither sine nor cosine is an algebraic function. Perhaps your instructions are a bit off.

You could just work a little magic:

\(\displaystyle \L\;sin(x) + cos(x) = \sqrt{2}\cos(x-\frac{\pi}{4})\)

That should make it a bit simpler.

 

[JamesMclean]

That seems to make it more difficult for me.... If i did do that, what would X =?

 

[tkhunny]

It's only one trig function. Definitely simpler.

Are you acquainted with the unit circle and various exact values?

What is the cosine of \(\displaystyle \frac{\pi}{6}\) or \(\displaystyle \frac{\pi}{4}\) or \(\displaystyle \frac{\pi}{3}\), for example?

 

[soroban]

Hello, James!

Solve for \(\displaystyle x\), where \(\displaystyle \,0\,\leq\,x\,\leq\,2\pi\)

\(\displaystyle \sin x \,+\,\cos x \;=\;1\)

Square both sides: \(\displaystyle \\sin x\,+\,\cos x)^2\;=\;1^2\)

We get: \(\displaystyle \:\sin^2x\,+\,2\cdot\sin x\cdot\cos x\,+\,\cos^2x\;=\;1\)

Since \(\displaystyle \sin^2x\,+\,\cos^2x\:=\:1\), we have: \(\displaystyle \:2\cdot\sin x\cdot\cos x\,+\,1\;=\;1\;\;\Rightarrow\;\;2\cdot\sin x\cdot\cos x\;=\;0\)

Since \(\displaystyle 2\cdot\sin x\cdot\cos x \:=\:\sin2x\), we have: \(\displaystyle \:\sin2x\;=\;0\)

. . Then: \(\displaystyle \:2x\;=\;0,\:\pi,\:3\pi,\:4\pi\)

. . Hence: \(\displaystyle \:x \;=\;0,\:\frac{\pi}{2},\:\pi,\:\frac{3\pi}{2}\)


Checking our answers, we find that the last two roots are extraneous.

Therefore: \(\displaystyle \;\fbox{x\;=\;0,\:\pi}\)

 

[JamesMclean]

thanks alot, That's a way that i actually understand it!

 

[tkhunny]

...but we'll never know if you have a clue about important things like simple, exact values around the unit circle. Oh, well.

 

[pka]

\(\displaystyle \pi\) is not a solution, but \(\displaystyle \frac{\pi} {2}\;\&\;2\pi\) are solutions.

 

[JamesMclean]

I know about all the exact values around the unit circle i just didn't realize what you were doing because i hadn't seen it done before. You don't want me to list them all out do you?

 

[Deleted member 4993]

Tkhunny was giving you hints to solve the problem in another way.


\(\displaystyle \L\;sin(x) + cos(x) = \sqrt{2}\cos(x-\frac{\pi}{4})\)

given

\(\displaystyle \L\;sin(x) + cos(x) = 1\)

so

\(\displaystyle \L\;\sqrt{2}\cos(x-\frac{\pi}{4}) = 1\)

\(\displaystyle \L\;cos(x-\frac{\pi}{4}) = \frac{\1}{\sqrt{2}\\)

\(\displaystyle \L\;cos(x-\frac{\pi}{4}) = cos(2*n*{pi} + \frac{\pi}{4}) , cos(2*n*{pi} - \frac{\pi}{4})\\)

\(\displaystyle \L\;x = 2*n*{pi} + \frac{\pi}{2} , 2*n*{pi}\\)

Now you can choose correct values of n within the given domain to solve for 'x'

 

[tkhunny]

I know about all the exact values around the unit circle i just didn't realize what you were doing because i hadn't seen it done before. You don't want me to list them all out do you?

Actually, it would help. You can talk your way through much of this one.

sin(x) + cos(x) = 1

Where is sin(x) = 1? \(\displaystyle x = \frac{\pi}{2}\) Is cos(\(\displaystyle \frac{\pi}{2}\)) = 0? Yes.

Where is cos(x) = 1? x = 0 Is sin(0) = 0? Yes.

You are well on your way to a complete solution. The basic values around the unit circle will help you if you review them.

 

[JamesMclean]

OK thanks tkhunny

I just hadn't seen the identity that you put up before.

Makes sense now with all the people talking about it.