c2 trigonometrical identities and simple equestions

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  1. reb0xx's Avatar
    1 07-04-2012  03:19
    • Exalted Member
    • Posts: 255
    c2 trigonometrical identities and simple equestions
    cos3q=-1 (q is the degree)
    the answers are 60, 180 and 300
    so i dont know how to solve this
    this what i did, the intervals 3*0 and 3*360=1080
    so my values are 180 , 360 , 540, 720, 900, 1080.
    cos3q=180 , 360 , 540, 720, 900, 1080
    (divide by 3)=60, 120, 180, 300, 360
  2. raheem94's Avatar
    2 07-04-2012  03:45
    • TSR Demigod
    • Posts: 5,512
    Re: c2 trigonometrical identities and simple equestions
    (Original post by reb0xx)
    cos3q=-1 (q is the degree)
    the answers are 60, 180 and 300
    so i dont know how to solve this
    this what i did, the intervals 3*0 and 3*360=1080
    so my values are 180 , 360 , 540, 720, 900, 1080.
    cos3q=180 , 360 , 540, 720, 900, 1080
    (divide by 3)=60, 120, 180, 300, 360
    \displaystyle cos3q=-1

    The interval is, \displaystyle 0\le q \le 360
    The interval for '3q' is \displaystyle 0\le 3q \le 1080

    Solving \displaystyle cos3q=-1 will give us 3 solutions. One solution you will get from the calculator the other two solutions can be found by adding 360 and 720 to the calculator value. Now divide the solutions by 3 to get the required values of 'q'.

    Below is the graph of y=cos(3q), this shows that it has 3 solution in the interval \displaystyle 0\le q \le 360 .

  3. CharlieBoardman's Avatar
    3 07-04-2012  03:48
    • Overlord in Training
    • Location: Manchester
    • Posts: 3,316
    Re: c2 trigonometrical identities and simple equestions
    (Original post by reb0xx)
    cos3q=-1 (q is the degree)
    the answers are 60, 180 and 300
    so i dont know how to solve this
    this what i did, the intervals 3*0 and 3*360=1080
    so my values are 180 , 360 , 540, 720, 900, 1080.
    cos3q=180 , 360 , 540, 720, 900, 1080
    (divide by 3)=60, 120, 180, 300, 360
    \cos 3\theta = -1 \cos ^-1 (-1) = 180

    For the next two values, add 360 each time, as the Cosine graph repeats every
    360^{\circ} We have 180, (180+360), (180+360+360) Which is; 180, 540, 900 So, 3\theta = 180, 540, 900 \theta = 60, 180, 300

    Does this help?
    Last edited by CharlieBoardman; 07-04-2012 at 03:49.
  4. reb0xx's Avatar
    4 07-04-2012  05:24
    • Exalted Member
    • Posts: 255
    Re: c2 trigonometrical identities and simple equestions
    (Original post by raheem94)
    \displaystyle cos3q=-1

    The interval is, \displaystyle 0\le q \le 360
    The interval for '3q' is \displaystyle 0\le 3q \le 1080

    Solving \displaystyle cos3q=-1 will give us 3 solutions. One solution you will get from the calculator the other two solutions can be found by adding 360 and 720 to the calculator value. Now divide the solutions by 3 to get the required values of 'q'.

    Below is the graph of y=cos(3q), this shows that it has 3 solution in the interval \displaystyle 0\le q \le 360 .

    thanks man i solved this
    but can you help me with this one
    4sinq=tanq
  5. raheem94's Avatar
    5 07-04-2012  05:49
    • TSR Demigod
    • Posts: 5,512
    Re: c2 trigonometrical identities and simple equestions
    (Original post by reb0xx)
    thanks man i solved this
    but can you help me with this one
    4sinq=tanq
    \displaystyle 4sinq = tanq

    Remember, \displaystyle tanq=\frac{sinq}{cosq}

    \displaystyle 4sinq = tanq \implies 4sinq=\frac{sinq}{cosq} \implies 4sinq - \frac{sinq}{cosq} =0

    Now factorise the above expression, e.g. \displaystyle 4x -\frac{x}{y}=0 \implies x\left(4-\frac1{y}\right)=0 This gives \displaystyle x=0 and \displaystyle 4-\frac1{y}\right = 0
  6. reb0xx's Avatar
    6 07-04-2012  06:14
    • Exalted Member
    • Posts: 255
    Re: c2 trigonometrical identities and simple equestions
    (Original post by raheem94)
    \displaystyle 4sinq = tanq

    Remember, \displaystyle tanq=\frac{sinq}{cosq}

    \displaystyle 4sinq = tanq \implies 4sinq=\frac{sinq}{cosq} \implies 4sinq - \frac{sinq}{cosq} =0

    Now factorise the above expression, e.g. \displaystyle 4x -\frac{x}{y}=0 \implies x\left(4-\frac1{y}\right)=0 This gives \displaystyle x=0 and \displaystyle 4-\frac1{y}\right = 0
    why did you chose sin to be x ?
  7. raheem94's Avatar
    7 07-04-2012  06:17
    • TSR Demigod
    • Posts: 5,512
    Re: c2 trigonometrical identities and simple equestions
    (Original post by reb0xx)
    why did you chose sin to be x ?
    I was giving you an example.
  8. reb0xx's Avatar
    8 07-04-2012  06:46
    • Exalted Member
    • Posts: 255
    Re: c2 trigonometrical identities and simple equestions
    (Original post by raheem94)
    I was giving you an example.
    ok one last question
    i can solve this equations using with the quadratic equation formula
    this one(http://www.oncalc.com/wp-content/upl...calculator.gif)
    are we allowed to solve this with a formula or by completing the square instead of factorizing it???
    are we allowed to do that in the exams?
  9. raheem94's Avatar
    9 07-04-2012  13:33
    • TSR Demigod
    • Posts: 5,512
    Re: c2 trigonometrical identities and simple equestions
    (Original post by reb0xx)
    ok one last question
    i can solve this equations using with the quadratic equation formula
    this one(http://www.oncalc.com/wp-content/upl...calculator.gif)
    are we allowed to solve this with a formula or by completing the square instead of factorizing it???
    are we allowed to do that in the exams?
    Which question are you talking about?

    If the question doesn't specifies a method then you can use any method.

    In your previous question, we got, \displaystyle 4sinq - \frac{sinq}{cosq} = 0

    Here there is no point of using any formula or completing the square, this can easily be solved by factorisation.

    \displaystyle 4sinq - \frac{sinq}{cosq} = 0 \implies sinq\left(4-\frac1{cosq}\right)=0

    So we get two equations, \displaystyle sinq=0 and \displaystyle 4-\frac1{cosq}=0

    Do you get it?
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