Results 1 to 6 of 6

Thread: easy trig question

  1. #1
    Raj
    Raj is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    60

    easy trig question

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

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1
    Quote Originally Posted by Raj View Post
    How would i solve questions such as sin(x)=$\displaystyle \frac{1}{2}$ or cos(3x)=$\displaystyle \frac{-\sqrt{3}}{2}$ without using a calculator.

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

    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 $\displaystyle \cos 3x=-\frac{\sqrt{3}}{2}$

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Raj
    Raj is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    60
    Quote Originally Posted by mathaddict View Post
    $\displaystyle \sin x =-\frac{1}{2}$

    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 $\displaystyle \cos 3x=-\frac{\sqrt{3}}{2}$

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1
    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 ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Raj
    Quote Originally Posted by Raj View Post
    How would i solve questions such as sin(x)=$\displaystyle \frac{1}{2}$ or cos(3x)=$\displaystyle \frac{-\sqrt{3}}{2}$ 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 $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$ and that $\displaystyle \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$.

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

    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 $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$ means that the sine of all these angles will also equal one-half: $\displaystyle \tfrac{5\pi}{6},\tfrac{13\pi}{6},\tfrac{17\pi}{6}, ...$

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

    Practice makes perfect!

    Grandad
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Raj
    Raj is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    60
    @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 $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$ and that $\displaystyle \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$.
    Thank-you, this clears things up

    Technique (check), experience (lacking)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quite an easy trig identity question - need pointers
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: Nov 14th 2010, 06:55 AM
  2. Easy trig question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Oct 28th 2010, 03:51 PM
  3. Pretty easy trig. question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jul 20th 2008, 08:22 PM
  4. Quick and easy question trig identities
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jan 12th 2008, 03:37 PM
  5. really easy trig question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 31st 2005, 03:37 AM

Search Tags


/mathhelpforum @mathhelpforum