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anonymous
 one year ago
5(cos^2)60 + 4(sec^2)45  (tan^2)45 = x
Find x
@Michele_Laina
anonymous
 one year ago
5(cos^2)60 + 4(sec^2)45  (tan^2)45 = x Find x @Michele_Laina

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0is 40 degrees the angle of the second term?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0we have to keep in mind these values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 45 = \frac{1}{{\cos 45}} = \frac{1}{{\sqrt 2 /2}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \hfill \\ {\left( {\sec 45} \right)^2} = 2 \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please substitute them into your equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OMG!!! ISY SEC 30 NETHER 40 NOR 45 :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0ok! I will rewrite the right value

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here are the right values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 30 = \frac{1}{{\cos 30}} = \frac{1}{{\sqrt 3 /2}} = \frac{2}{{\sqrt 3 }} \hfill \\ {\left( {\sec 30} \right)^2} = \frac{4}{3} \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0so, after a simple substitution, we get: \[\Large \left( {5 \times \frac{1}{4}} \right) + \left( {4 \times \frac{4}{3}} \right)  1 = x\]
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