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anonymous

  • one year ago

cot x sec4x = cot x + 2 tan x + tan3x

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  1. anonymous
    • one year ago
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    Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

  2. Michele_Laino
    • one year ago
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    I think that we have to use these identities: \[\Large \begin{gathered} {\left( {\sec x} \right)^4} = \frac{1}{{{{\left( {\cos x} \right)}^4}}} \hfill \\ \hfill \\ {\left( {\tan x} \right)^3} = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} \hfill \\ \end{gathered} \]

  3. anonymous
    • one year ago
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    ok

  4. Michele_Laino
    • one year ago
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    so, left side is: \[\Large \cot x{\left( {\sec x} \right)^4} = \frac{{\cos x}}{{\sin x}}\frac{1}{{{{\left( {\cos x} \right)}^4}}} = ...?\] please simplify

  5. anonymous
    • one year ago
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    \[\cos x/\sin x \times \left( cosx \right)^4\]

  6. Michele_Laino
    • one year ago
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    please you have to simplify cos(x) with (cos x)^4

  7. anonymous
    • one year ago
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    1/sin x * (cosx)^3

  8. Michele_Laino
    • one year ago
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    ok! correct!

  9. Michele_Laino
    • one year ago
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    now, we have to make the right side, equal to your expression for left side

  10. Michele_Laino
    • one year ago
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    so, we can write: \[\large \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3}\]

  11. Michele_Laino
    • one year ago
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    or: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    what is the common denominator?

  13. anonymous
    • one year ago
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    sinx*cosx^3

  14. Michele_Laino
    • one year ago
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    correct!

  15. Michele_Laino
    • one year ago
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    so, we can write this: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\ \end{gathered} \]

  16. anonymous
    • one year ago
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    ok I get it

  17. Michele_Laino
    • one year ago
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    we can simplify like this: \[\large \begin{gathered} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \end{gathered} \]

  18. anonymous
    • one year ago
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    \[(\cos(x)^4 +2\sin(x)^2(cosx)^2+\sin(x)^4)/\sin(x)xcos(x)^3\]

  19. Michele_Laino
    • one year ago
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    hint: the numerator is a perfect square

  20. Michele_Laino
    • one year ago
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    hint: \[\Large {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = ...?\]

  21. anonymous
    • one year ago
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    (sinx)^4+2(sinx)^2(cosx)^2+(cosx)^2

  22. anonymous
    • one year ago
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    (cosx)^4

  23. anonymous
    • one year ago
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    ((sinx^2 + cosx^2)^2)/sinx(cosx)^3

  24. Michele_Laino
    • one year ago
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    I got this: since \[\Large \begin{gathered} {\left( {\cos x} \right)^4} + 2{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^2} + \left( {\sin x} \right) = \hfill \\ = {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = {1^2} = 1 \hfill \\ \end{gathered} \] we have: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{1}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\ \hfill \\ \end{gathered} \]

  25. anonymous
    • one year ago
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    oh

  26. anonymous
    • one year ago
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    ok I understand thanks

  27. Michele_Laino
    • one year ago
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    sorry, another typo... \[\Large \begin{gathered} {\left( {\cos x} \right)^4} + 2{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^2} + {\left( {\sin x} \right)^4} = \hfill \\ = {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = {1^2} = 1 \hfill \\ \end{gathered} \]

  28. anonymous
    • one year ago
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    ok thanks

  29. Michele_Laino
    • one year ago
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    :)

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