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Math Help - Need some help with arcus functions

  1. #1
    Vey
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    Need some help with arcus functions

    Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

    My problem is this:
    I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Vey View Post
    Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

    My problem is this:
    I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

    Thanks in advance!
    let a = \arcsin\left(\frac{3}{\sqrt{10}}\right)

    b = \arccos\left(\frac{1}{\sqrt{5}}\right)

    \sin(a+b) = \sin{a}\cos{b} + \cos{a}\sin{b}

    \sin(a+b) = \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{2}}<br />

    a+b = \frac{3\pi}{4}
    Last edited by skeeter; January 2nd 2010 at 08:54 AM. Reason: angle is in quad II ... mea culpa.
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  3. #3
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    Hello Vey

    Welcome to Math Help Forum!
    Quote Originally Posted by Vey View Post
    Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

    My problem is this:
    I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

    Thanks in advance!
    The best thing to do with inverse trig functions is to get rid of them as soon as you can! So:

    Let:
    x = \arcsin \left(\frac{3}{\sqrt{10}}\right)
    and
    y = \arccos \left(\frac{1}{\sqrt5}\right)
    Then get rid of the inverse functions:
    \sin x = \frac{3}{\sqrt{10}} and \cos y =  \frac{1}{\sqrt5}
    So, using Pythagoras' Theorem:
    \cos x = \frac{1}{\sqrt{10}} and \sin y = \frac{2}{\sqrt5}
    Now we need x + y, so, using \cos(x+y) = \cos x \cos y - \sin x \sin y, we get:
    \cos(x+y) = \frac{1}{\sqrt{10}}\frac{1}{\sqrt{5}}- \frac{3}{\sqrt{10}}\frac{2}{\sqrt{5}}
    =\frac{1}{5\sqrt2}-\frac{6}{5\sqrt2}

    = -\frac{1}{\sqrt2}
    Of course, there's an infinite number of angles whose cosine is = -\frac{1}{\sqrt2}, but, noting that 0<x<\frac{\pi}{2} and 0 < y < \frac{\pi}{2}, it's clear that:
    x+y = \frac{3\pi}{4}
    Grandad
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  4. #4
    Vey
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    Thanks guys, the second method proved useful!
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