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anonymous

  • 5 years ago

the value of tan(2tan^-1 (1/5)-pi/4) is?

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  1. anonymous
    • 5 years ago
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    Let\[\theta=2\tan^{-1} (1/5)-(\pi/4)\]Form a triangle with tan theta

  2. anonymous
    • 5 years ago
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    how?

  3. anonymous
    • 5 years ago
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    How what?!

  4. anonymous
    • 5 years ago
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    wrong!

  5. anonymous
    • 5 years ago
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    let x= \[\tan^{-1} \frac{1}{5}\]

  6. anonymous
    • 5 years ago
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    so our expression becomes \[\tan(2x-\frac{\pi}{4})\]

  7. anonymous
    • 5 years ago
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    now apply difference of angle formula

  8. anonymous
    • 5 years ago
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    \[\tan(A-B) = \frac{\tan(A)-\tan(B) } {1+\tan(A)\tan(B) } \]

  9. anonymous
    • 5 years ago
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    so we get

  10. anonymous
    • 5 years ago
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    \[\frac{\tan(2x) -1}{1+\tan(2x)} \]

  11. anonymous
    • 5 years ago
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    but ,\[\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)} \]

  12. anonymous
    • 5 years ago
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    now, we go back to our substitution x= tan^-1(1/5) take tan of both sides , and we get tan(x) = 1/5 now we can draw up a general right angle triangle and mark an angle x , and fill in the lengths 1 and 5 for thre opposite and adjacent sides respectively remember tan = opposite/adjacent

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  13. anonymous
    • 5 years ago
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    note the hypotenuse is sqrt(26)

  14. anonymous
    • 5 years ago
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    now from that triangle we can find the value of sin(x) and cos(x) , the then you just plug them into the expression we had above

  15. anonymous
    • 5 years ago
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    so \[\sin(x) = \frac{1}{\sqrt{26}}\] \[\cos(x) = \frac{5}{\sqrt{26}}\] just by using the definitions

  16. anonymous
    • 5 years ago
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    so , our whole expression was\[\tan(2x- \frac{\pi}{4}) = \frac{ \frac{ 2\sin(x)\cos(x) }{\cos^2(x)-\sin^2(x) } -1}{ 1+ \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}\]

  17. anonymous
    • 5 years ago
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    then sub in the values for sin(x) and cos(x) above, and simplify

  18. anonymous
    • 5 years ago
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    you can do that

  19. anonymous
    • 5 years ago
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    good. it worked out well. thanx man!

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