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Math Help - more inverse trigonometric functions

  1. #1
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    more inverse trigonometric functions

    a) show that tan-1 (4) - tan-1(3/5) =pi/4
    b) tan-1(5/12)+cos-1(5/13)=pi/2
    okayy i dont understand inverse trig. i draw the diagrams and i input it into my calculator but they are never exact values so how do you set it out and find it showing the steps to get there and not just putting the whole thing into the calculator?

    thankyou to anyone who can help me !
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  2. #2
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    Quote Originally Posted by iiharthero View Post
    a) show that tan-1 (4) - tan-1(3/5) =pi/4
    b) tan-1(5/12)+cos-1(5/13)=pi/2
    let a = \arctan(4) , b = \arctan(3/5)

    that means \tan(a) = 4 and \tan(b) = \frac{3}{5}

    \tan(a-b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} =

     <br />
\frac{4 - \frac{3}{5}}{1 + 4\left(\frac{3}{5}\right)} =<br />

     <br />
\frac{20-3}{5 + 12} = 1<br />

    so, \tan(a-b) = 1

     <br />
\arctan(1) = a - b = \frac{\pi}{4}<br />
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