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Math Help - trigo solving

  1. #1
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    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 ??
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  2. #2
    MHF Contributor red_dog's Avatar
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    You have just two cases:

    \sin 2x=\frac{1}{2} and \sin 2x=-\frac{1}{2}
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  3. #3
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    Quote Originally Posted by red_dog View Post
    You have just two cases:

    \sin 2x=\frac{1}{2} and \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 .
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  4. #4
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    Hello thereddevils

    As red_dog has said, you need to solve two equations

    \sin2x = \frac12

    and

    \sin2x=-\frac12

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

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

    -\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 2x. Divide them all by 2, and you're done.

    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello thereddevils

    As red_dog has said, you need to solve two equations

    \sin2x = \frac12

    and

    \sin2x=-\frac12

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

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

    -\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 2x. Divide them all by 2, and you're done.

    Grandad
    Thanks Grandad ,

    sin 2x =1/2

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

    sin 2x = -1/2

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

    How to get the negative one ??

    Thanks .
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    How to get the negative one ??

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

    Grandad
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  7. #7
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    Quote Originally Posted by Grandad View Post
    Use the rotational symmetry of the graph. The negative part is just a 180^o rotation of the positive part. So where you have \pi/6 on the right-hand side, you'll have -\pi/6 on the left.

    Grandad
    THanks Grandad , does this also work for cos and tan ?
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    THanks Grandad , does this also work for cos and tan ?
    Absolutely! Just sketch the graph, then use the section between 0 and \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 0 to \pi/2, you can easily work out the corresponding angle in any other range.

    Grandad
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  9. #9
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    Quote Originally Posted by Grandad View Post
    Hello thereddevilsAbsolutely! Just sketch the graph, then use the section between 0 and \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 0 to \pi/2, you can easily work out the corresponding angle in any other range.

    Grandad
    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 ..
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  10. #10
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    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 ..
    Have a look at the diagram I've attached, which shows a sketch-graph of y = \tan x, between \pm360^o, and the line y = 1 as a dotted line.

    You'll see that I've drawn a rectangle around the section of the graph between x = 0 and x = 90^o - rectangle number (1). This rectangle is then rotated about the point (90,0) to form rectangle (2). And it has been translated through 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 \tan x = 1 is x = 45^o. Using this graph you can see that there are others where x = 180 +45 and -180+45, -360 + 45, ... So the solutions between \pm180^o are -135^o, 45^o.

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

    Grandad
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