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anonymous

  • one year ago

5(cos^2)60 + 4(sec^2)45 - (tan^2)45 = x Find x @Michele_Laina

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  1. Michele_Laino
    • one year ago
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    is 40 degrees the angle of the second term?

  2. anonymous
    • one year ago
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    45*

  3. Michele_Laino
    • one year ago
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    ok!

  4. anonymous
    • one year ago
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    ? How to solve? ?

  5. Michele_Laino
    • one year ago
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    we have to keep in mind these values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 45 = \frac{1}{{\cos 45}} = \frac{1}{{\sqrt 2 /2}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \hfill \\ {\left( {\sec 45} \right)^2} = 2 \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]

  6. anonymous
    • one year ago
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    Ok!

  7. Michele_Laino
    • one year ago
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    please substitute them into your equation

  8. anonymous
    • one year ago
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    OMG!!! ISY SEC 30 NETHER 40 NOR 45 :(

  9. Michele_Laino
    • one year ago
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    ok! I will rewrite the right value

  10. anonymous
    • one year ago
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    OK!

  11. Michele_Laino
    • one year ago
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    here are the right values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 30 = \frac{1}{{\cos 30}} = \frac{1}{{\sqrt 3 /2}} = \frac{2}{{\sqrt 3 }} \hfill \\ {\left( {\sec 30} \right)^2} = \frac{4}{3} \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    so, after a simple substitution, we get: \[\Large \left( {5 \times \frac{1}{4}} \right) + \left( {4 \times \frac{4}{3}} \right) - 1 = x\]

  13. Michele_Laino
    • one year ago
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    please simplify

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