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Math Help - easy trig question

  1. #1
    Raj
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    easy trig question

    How would i solve questions such as sin(x)= \frac{1}{2} or cos(3x)= \frac{-\sqrt{3}}{2} without using a calculator.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Raj View Post
    How would i solve questions such as sin(x)= \frac{1}{2} or cos(3x)= \frac{-\sqrt{3}}{2} without using a calculator.

    Thanks in advance
    \sin x =-\frac{1}{2}

    first of all , the reference angle is 30 degrees . But sin is negative so it would be in the 3rd and 4th quadrant ie

    x=210 , 330

    for \cos 3x=-\frac{\sqrt{3}}{2}

    reference angle = 30 degrees and since cos is negative , it will be in the 2nd and 3rd quadrant .

    ie 3x=150 , 210 , 510 , 570 , 870 , 930

    x=50 , 70 , 170 , 190 , 290 , 310

    Assuming that 0<x<360 , then 0<3x<1080
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  3. #3
    Raj
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    Quote Originally Posted by mathaddict View Post
    \sin x =-\frac{1}{2}

    first of all , the reference angle is 30 degrees . But sin is negative so it would be in the 3rd and 4th quadrant ie

    x=210 , 330

    for \cos 3x=-\frac{\sqrt{3}}{2}

    reference angle = 30 degrees and since cos is negative , it will be in the 2nd and 3rd quadrant .

    ie 3x=150 , 210 , 510 , 570 , 870 , 930

    x=50 , 70 , 170 , 190 , 290 , 310

    Assuming that 0<x<360 , then 0<3x<1080
    thanks but i really have no idea how you got the reference angle
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  4. #4
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    Quote Originally Posted by Raj View Post
    thanks but i really have no idea how you got the reference angle
    lets say for sin x = -1/2

    we just ignore the negative sign first , ie sin x =1/2

    So by recalling these special angles , x=30 degree , then from here we only the negative sign by deciding which quadrant should it be in .

    Another example ,

    cos x = - 0.445

    same thing , ignore the sign first , then use the calculator to find the reference angle (63.58) , then now since its negative so cos will be in 2nd and 3rd quadrant ..

    Clear ?
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  5. #5
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    Hello Raj
    Quote Originally Posted by Raj View Post
    How would i solve questions such as sin(x)= \frac{1}{2} or cos(3x)= \frac{-\sqrt{3}}{2} without using a calculator.

    Thanks in advance
    The answer is: using a combination of experience - you simply recognise certain trig ratios - and technique.

    For instance, you'll need experience - and the ability to memorise certain facts - to know that \sin\tfrac{\pi}{6} = \tfrac12 and that \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}.

    Then you need a technique that will enable you to handle negative signs: for instance, how to use the fact that \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2} to enable you to find an angle whose cosine is -\tfrac{\sqrt3}{2}. (One such angle is \tfrac{5\pi}{6}.)

    Then you'll need a technique that will enable you to find other angles with the same sine or cosine. For example, the fact that \sin\tfrac{\pi}{6} = \tfrac12 means that the sine of all these angles will also equal one-half: \tfrac{5\pi}{6},\tfrac{13\pi}{6},\tfrac{17\pi}{6},  ...

    Finally, you'll need a technique that will enable you to deal with mutliple angles. For example, if \sin3x = \tfrac12, we've just seen that the possible values of 3x are \tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},\t  frac{17\pi}{6},.... So we'd divide each of these by 3 to get the values of x: \tfrac{\pi}{18},\tfrac{5\pi}{18},\tfrac{13\pi}{18}  ,\tfrac{17\pi}{18},...

    Practice makes perfect!

    Grandad
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  6. #6
    Raj
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    @mathaddict

    Ah special angles, I totally forgot. Thanks.

    Quote Originally Posted by Grandad View Post
    Hello RajThe answer is: using a combination of experience - you simply recognise certain trig ratios - and technique.

    For instance, you'll need experience - and the ability to memorise certain facts - to know that \sin\tfrac{\pi}{6} = \tfrac12 and that \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}.
    Thank-you, this clears things up

    Technique (check), experience (lacking)
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