anonymous
  • anonymous
cot x sec4x = cot x + 2 tan x + tan3x
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
Michele_Laino
  • Michele_Laino
I think that we have to use these identities: \[\Large \begin{gathered} {\left( {\sec x} \right)^4} = \frac{1}{{{{\left( {\cos x} \right)}^4}}} \hfill \\ \hfill \\ {\left( {\tan x} \right)^3} = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
ok

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Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
\[\cos x/\sin x \times \left( cosx \right)^4\]
Michele_Laino
  • Michele_Laino
please you have to simplify cos(x) with (cos x)^4
anonymous
  • anonymous
1/sin x * (cosx)^3
Michele_Laino
  • Michele_Laino
ok! correct!
Michele_Laino
  • Michele_Laino
now, we have to make the right side, equal to your expression for left side
Michele_Laino
  • Michele_Laino
so, we can write: \[\large \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3}\]
Michele_Laino
  • Michele_Laino
or: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
what is the common denominator?
anonymous
  • anonymous
sinx*cosx^3
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
so, we can write this: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
ok I get it
Michele_Laino
  • Michele_Laino
we can simplify like this: \[\large \begin{gathered} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \end{gathered} \]
anonymous
  • anonymous
\[(\cos(x)^4 +2\sin(x)^2(cosx)^2+\sin(x)^4)/\sin(x)xcos(x)^3\]
Michele_Laino
  • Michele_Laino
hint: the numerator is a perfect square
Michele_Laino
  • Michele_Laino
hint: \[\Large {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = ...?\]
anonymous
  • anonymous
(sinx)^4+2(sinx)^2(cosx)^2+(cosx)^2
anonymous
  • anonymous
(cosx)^4
anonymous
  • anonymous
((sinx^2 + cosx^2)^2)/sinx(cosx)^3
Michele_Laino
  • Michele_Laino
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: \[\large \begin{gathered} \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\ \hfill \\ = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\ \hfill \\ = \frac{1}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\ \hfill \\ \end{gathered} \]
anonymous
  • anonymous
oh
anonymous
  • anonymous
ok I understand thanks
Michele_Laino
  • Michele_Laino
sorry, another typo... \[\Large \begin{gathered} {\left( {\cos x} \right)^4} + 2{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^2} + {\left( {\sin x} \right)^4} = \hfill \\ = {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = {1^2} = 1 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
ok thanks
Michele_Laino
  • Michele_Laino
:)

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