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anonymous

  • one year ago

cos(arctan(-12/5)+arctan(3/4))

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  1. anonymous
    • one year ago
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    @dan815 @uri

  2. anonymous
    • one year ago
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    I'm assuming that you want to solve this in exact form by hand without a calculator.

  3. anonymous
    • one year ago
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    In that case, you will need to use the arctan sum formula.

  4. anonymous
    • one year ago
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    @math1234 that would be correct

  5. anonymous
    • one year ago
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    1/1+x^2 ?

  6. anonymous
    • one year ago
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    No, it is \[\tan^{-1} a + \tan^{-1} b = \tan^{-1} \frac{ a+b }{ 1-ab }\]

  7. anonymous
    • one year ago
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    Upon combining the inside using the arctan sum formula, you can use your mentioned formula to compute the cos of the arctan.

  8. anonymous
    • one year ago
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    so that gves us \[\tan^{-1} \frac{ \frac{ -12 }{ 5 }+\frac{ 3 }{ 4 } }{ 1-\frac{ -12 }{ 4 }*\frac{ 3 }{ 4 } }\]

  9. anonymous
    • one year ago
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    Yes, then you plug it into \[\cos (\tan^{-1} x) = \frac{ 1 }{ \sqrt{1+x^2} }\]

  10. anonymous
    • one year ago
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    Where x is your fractional expression above.

  11. anonymous
    • one year ago
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    \[\cos (\tan^{-1} \frac{ 33 }{ 16 })=\frac{ 1 }{ \sqrt{1+(\frac{ 33 }{ 6 }})^{2} }\]

  12. anonymous
    • one year ago
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    idk where to go from here

  13. anonymous
    • one year ago
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    That's your answer.

  14. anonymous
    • one year ago
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    Just add the denominator.

  15. anonymous
    • one year ago
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    56/65 Refer to the attachment below.

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