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Trigonometric ratio question

  • 29-09-2011 2:22pm
    #1
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    I've copied the question below from a website:
    In each question below you are given the quadrant an angle lies in and the value of one trigonometric ratio. In each case you should find the value of a stated second trigonometric ratio without finding the angle itself (i.e. without using inverse trigonometric functions).
    Give answers to 3 decimal places.
    (a) 0 < x < 90o, sin x = 0.3, what is cos x ?
    Now, the way I would have immediately answered this is to use inverse trigonometric ratio and find the angle. So I was wondering what method would be used since this is ruled out in the question?

    Thank you


Comments

  • Registered Users, Registered Users 2 Posts: 360 ✭✭CJC86


    In these sort of questions you should have a quick look at your log tables and see what equations include [latex]\cos x[/latex] and [latex]\sin x[/latex].

    Here you would use [latex] \cos^2 x + \sin^2 x = 1[/latex]. Then you work out whether [latex]\cos x[/latex] should be positive or negative, depending on which quadrant x is in.


  • Registered Users, Registered Users 2 Posts: 434 ✭✭Smythe


    Thank you. I'll use trig identities. Thanks again.


  • Registered Users, Registered Users 2 Posts: 360 ✭✭CJC86


    Forgot to mention the easy way to do this, or should I say the more fundamental way.

    If sinx = 0.3, then we can model this with a right angled triangle with one side of length 3 and hypothenuse of length 10, then use Pythagoras' Theorem to get the length of the other side. Then you can quickly work out cosx and tanx, for x between 0 and 90 degrees. Then work out whether cosx should be positive or negative depending on which quadrant x is in.


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