trigo solving

• Oct 2nd 2009, 04:09 AM
thereddevils
trigo solving
Find the value of x of the following trigonometric equation with x is between -pi and pi

| sin 2x | = 1/2

My work

sin 2x = 1/2 (case 1)

2x= 30 , 150

x= 15 and 75

sin (-2x) = 1/2

There is no solution for this .

case 2

sin 2x = -1/2

no solution

sin (-2x) = -1/2

-2x = 30 , 150

x = -15 , -75

Am i correct ??
• Oct 2nd 2009, 07:14 AM
red_dog
You have just two cases:

$\displaystyle \sin 2x=\frac{1}{2}$ and $\displaystyle \sin 2x=-\frac{1}{2}$
• Oct 2nd 2009, 07:24 AM
thereddevils
Quote:

Originally Posted by red_dog
You have just two cases:

$\displaystyle \sin 2x=\frac{1}{2}$ and $\displaystyle \sin 2x=-\frac{1}{2}$

so the solution will be just pi/4 and 5pi/12 but this is for the range 0 and pi only . how about -pi and 0 ?

THanks .
• Oct 2nd 2009, 08:04 AM
Hello thereddevils

As red_dog has said, you need to solve two equations

$\displaystyle \sin2x = \frac12$

and

$\displaystyle \sin2x=-\frac12$

Since you want values of $\displaystyle x$ between $\displaystyle -\pi$ and $\displaystyle \pi$, you'll need to look at values of $\displaystyle 2x$ between $\displaystyle -2\pi$ and $\displaystyle 2\pi$.

I always find the graph of $\displaystyle \sin\theta$ easier to use than the unit circle, so look at the sketch I've attached. The easiest value is, of course, $\displaystyle \sin(\tfrac{\pi}{6})=\tfrac12$. Look at where the dotted lines intersect the graph. They are at:

$\displaystyle -\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7 \pi}{6},\frac{11\pi}{6}$

So these are the possible values of $\displaystyle 2x$. Divide them all by $\displaystyle 2$, and you're done.

• Oct 2nd 2009, 08:30 AM
thereddevils
Quote:

Originally Posted by Grandad
Hello thereddevils

As red_dog has said, you need to solve two equations

$\displaystyle \sin2x = \frac12$

and

$\displaystyle \sin2x=-\frac12$

Since you want values of $\displaystyle x$ between $\displaystyle -\pi$ and $\displaystyle \pi$, you'll need to look at values of $\displaystyle 2x$ between $\displaystyle -2\pi$ and $\displaystyle 2\pi$.

I always find the graph of $\displaystyle \sin\theta$ easier to use than the unit circle, so look at the sketch I've attached. The easiest value is, of course, $\displaystyle \sin(\tfrac{\pi}{6})=\tfrac12$. Look at where the dotted lines intersect the graph. They are at:

$\displaystyle -\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7 \pi}{6},\frac{11\pi}{6}$

So these are the possible values of $\displaystyle 2x$. Divide them all by $\displaystyle 2$, and you're done.

sin 2x =1/2

so 2x = $\displaystyle \frac{\pi}{6},\frac{5\pi}{6}$

sin 2x = -1/2

$\displaystyle 2x= \frac{7\pi}{6},\frac{11\pi}{6}$

How to get the negative one ??

Thanks .
• Oct 2nd 2009, 12:26 PM
Quote:

Originally Posted by thereddevils
How to get the negative one ??

Thanks .

Use the rotational symmetry of the graph. The negative part is just a $\displaystyle 180^o$ rotation of the positive part. So where you have $\displaystyle \pi/6$ on the right-hand side, you'll have $\displaystyle -\pi/6$ on the left.

• Oct 2nd 2009, 07:00 PM
thereddevils
Quote:

Originally Posted by Grandad
Use the rotational symmetry of the graph. The negative part is just a $\displaystyle 180^o$ rotation of the positive part. So where you have $\displaystyle \pi/6$ on the right-hand side, you'll have $\displaystyle -\pi/6$ on the left.

THanks Grandad , does this also work for cos and tan ?
• Oct 2nd 2009, 09:55 PM
Hello thereddevils
Quote:

Originally Posted by thereddevils
THanks Grandad , does this also work for cos and tan ?

Absolutely! Just sketch the graph, then use the section between $\displaystyle 0$ and $\displaystyle \pi/2$ (the bit I've indicated below inside the red rectangle) as a 'building block', rotating and reflecting it as necessary, to create the whole graph, for any range of values you want. Then, provided you know the angle in the range $\displaystyle 0$ to $\displaystyle \pi/2$, you can easily work out the corresponding angle in any other range.

• Oct 3rd 2009, 02:37 AM
thereddevils
Quote:

Originally Posted by Grandad
Hello thereddevilsAbsolutely! Just sketch the graph, then use the section between $\displaystyle 0$ and $\displaystyle \pi/2$ (the bit I've indicated below inside the red rectangle) as a 'building block', rotating and reflecting it as necessary, to create the whole graph, for any range of values you want. Then, provided you know the angle in the range $\displaystyle 0$ to $\displaystyle \pi/2$, you can easily work out the corresponding angle in any other range.

THanks Grandad , i see . Can i bring in another question which i asked in another thread for reference .
Solve for x for the range of x between -pi and pi
sin x = cos x

tan x=1

x=45

Would the other x be -45 ?? I don think so .. (Thinking)
• Oct 3rd 2009, 07:27 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
THanks Grandad , i see . Can i bring in another question which i asked in another thread for reference .
Solve for x for the range of x between -pi and pi
sin x = cos x

tan x=1

x=45

Would the other x be -45 ?? I don think so .. (Thinking)

Have a look at the diagram I've attached, which shows a sketch-graph of $\displaystyle y = \tan x$, between $\displaystyle \pm360^o$, and the line $\displaystyle y = 1$ as a dotted line.

You'll see that I've drawn a rectangle around the section of the graph between $\displaystyle x = 0$ and $\displaystyle x = 90^o$ - rectangle number (1). This rectangle is then rotated about the point $\displaystyle (90,0)$ to form rectangle (2). And it has been translated through $\displaystyle 180^o$ to form rectangle (3) ... And so on, to create as many sections of the graph as you need.

Now we know that the first value where $\displaystyle \tan x = 1$ is $\displaystyle x = 45^o$. Using this graph you can see that there are others where $\displaystyle x = 180 +45$ and $\displaystyle -180+45$, $\displaystyle -360 + 45$, ... So the solutions between $\displaystyle \pm180^o$ are $\displaystyle -135^o, 45^o$.

We can also work out in the same way where $\displaystyle \tan x = -1$. Rectangle (2) shows that there will be one value at $\displaystyle x = 180-45 = 135^o$. And there'll be others at $\displaystyle x = 360-45 = 315^o,-45^o, -180-45 = -225^o, ...$etc.

Provided you can sketch the graphs of sine, cosine and tangent quickly - and it's worth learning them - I think you'll find this is a great help in solving trig equations.