trig functions and exact values

**Stevo_Evo_22** · July 1st 2008, 08:13 PM

OK so the other day Reckoner gave me some great help with these, but now i can't work out this one. What's really confused me is the $x-pi/3$ in the middle. I know the answer should be over 6 (as in like $pi/6$ ) but I can't work out how the bit in the middle comes into play. Would you do this...

$sinx-sin*pi/3=1/2<br /> sinx=1/2 + sin*3<br />$
etc.....????????????????????

**Mathstud28** · July 1st 2008, 08:15 PM

Originally Posted by Stevo_Evo_22

OK so the other day Reckoner gave me some great help with these, but now i can't work out this one. What's really confused me is the $x-pi/3$ in the middle. I know the answer should be over 6 (as in like $pi/6$ ) but I can't work out how the bit in the middle comes into play. Would you do this...

$sinx-sin*pi/3=1/2<br /> sinx=1/2 + sin*3<br />$
etc.....????????????????????

I have, not even in the slightest, any idea what your question is.

**Stevo_Evo_22** · July 1st 2008, 08:17 PM

lol sorry ok....

If 0 is the same as or less than x, which is the same as or less than 2pi, find:

{x:sin(x-pi/3)=1/2}

**o_O** · July 1st 2008, 08:24 PM

Let $\theta = x - \frac{\pi}{3}$ for visualization purposes. Then:
$\sin \theta = \frac{1}{2} \: \: \Rightarrow \: \: \theta =\frac{\pi}{6}, \frac{5\pi}{6}$ (you should be familiar with these values)

Now, since $\theta = x - \frac{\pi}{3}$ , simply solve for x: $x - \frac{\pi}{3} = \frac{\pi}{6} \qquad x - \frac{\pi}{3} = \frac{5\pi}{6}$

**Mathstud28** · July 1st 2008, 08:26 PM

Originally Posted by Stevo_Evo_22

lol sorry ok....

If 0 is the same as or less than x, which is the same as or less than 2pi, find:

{x:sin(x-pi/3)=1/2}

So your saying solve

$\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}$

With the given condition that $x\leq{0}$ ?

If so,

$x-\frac{\pi}{3}=\frac{\pi}{6}+2\pi{n}\quad\text{wher e }n\in\mathbb{Z}$

So then we have that

$x=\frac{\pi}{2}+2\pi{n}\quad\text{where }n\in\mathbb{Z}$

But $x<0\Rightarrow\frac{\pi}{2}+2\pi{n}$

So this implies that $x=\frac{\pi}{2}+2\pi{n}\quad\text{where }n\in\mathbb{Z^-}$

**Stevo_Evo_22** · July 1st 2008, 08:28 PM

Perfect, thanks!

I was getting a similar answer, but didn't know what to do after i got 5pi/6.

Thanks a million and sorry about all these trig questions but we're on holidays so we can't get help from a teacher (and everyone here is always so helpful)

**Stevo_Evo_22** · July 1st 2008, 08:32 PM

[quote=Mathstud28;163443]So your saying solve

$\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}$

With the given condition that $x\leq{0}$ ?[quote]

No, where x is between or the same as 0 and 2pi (in a unit circle).

**Mathstud28** · July 1st 2008, 08:39 PM

[quote=Stevo_Evo_22;163445][quote=Mathstud28;163443]So your saying solve

$\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}$

With the given condition that $x\leq{0}$ ?

No, where x is between or the same as 0 and 2pi (in a unit circle).

oh

...sorry...a little communication problem

**Stevo_Evo_22** · July 1st 2008, 08:50 PM

Thats ok, dw bout it. Here's another one that doesn't really make sense:

x is between 0 and 2pi again but find: {x: cos(x-pi/6) = sqrt3/2}

Should the answer be 11pi/6 ?

**Simplicity** · July 2nd 2008, 03:27 AM

Originally Posted by Stevo_Evo_22

Thats ok, dw bout it. Here's another one that doesn't really make sense:

x is between 0 and 2pi again but find: {x: cos(x-pi/6) = sqrt3/2}

Should the answer be 11pi/6 ?

$\cos \left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \ \ \ \ 0 \le x \le 2\pi$

$\cos ^{-1} \left( \frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}, \frac{11\pi}{6}$

But these value satify the equation when $x-6 = \cos \left(x - \frac{\pi}{6}\right)$ and we want the value for $x$ hence add $6$ to both sides.

$\therefore x= \frac{\pi}{3}, 2\pi$

**Stevo_Evo_22** · July 2nd 2008, 04:55 AM

That's the answer i got, but the book's answer said it was wrong! I'm really getting sick of the books wrong answers

Thanks heaps!

**Simplicity** · July 2nd 2008, 04:58 AM

Originally Posted by Stevo_Evo_22

That's the answer i got, but the book's answer said it was wrong! I'm really getting sick of the books wrong answers

Thanks heaps!

If you are unsure then just insert your $x$ values back into the equation and see if you get the answer you require.

$x= \frac{\pi}{3} \implies \cos \left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ Therefore this value is correct.

$x= \frac{\pi}{3} \implies \cos \left({2\pi} - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ Therefore this value is correct too.

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