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Show the equation don't depend to x E = ((sin x)^2-(sin x)^4)/((cos x)^2)+1/(1+(tgx)^2)

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greenbel | Honors

Posted July 7, 2013 at 4:46 PM via web

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Show the equation don't depend to x

E = ((sin x)^2-(sin x)^4)/((cos x)^2)+1/(1+(tgx)^2)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 7, 2013 at 5:13 PM (Answer #1)

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You need to use the following trigonometric identity, such that:

`tan x = (sin x)/(cos x) => tan^2 x = (sin^2 x)/(cos^2 x)`

 

Replacing `(sin^2 x)/(cos^2 x)` for `tan^2 x` in the given expression, yields:

`E = (sin^2 x - sin^4 x)/(cos^2 x) + 1/(1 + (sin^2 x)/(cos^2 x))`

Bringing the terms in summation `1 + (sin^2 x)/(cos^2 x)` to a common denominator, yields:

`1 + (sin^2 x)/(cos^2 x) = (cos^2 x + sin^2 x)/(cos^2 x)`

Using the basic formula of trigonometry `cos^2 x + sin^2 x = 1 ` yields:

`1 + (sin^2 x)/(cos^2 x) = 1/(cos^2 x)`

Replacing `1/(cos^2 x)` for `1 + (sin^2 x)/(cos^2 x)` yields:

`E = (sin^2 x - sin^4 x)/(cos^2 x) + 1/(1/(cos^2 x))`

`E = (sin^2 x - sin^4 x)/(cos^2 x) + cos^2 x`

Bringing the terms to a common denominator yields:

`E = (sin^2 x - sin^4 x + cos^4 x)/(cos^2 x)`

You need to convert the difference of squares `cos^4 x - sin^4 x` into a product, such that:

`cos^4 x - sin^4 x= (cos^2 x - sin^2 x)(cos^2 x + sin^2 x)`

Since `cos^2 x + sin^2 x = 1` yields:

`cos^4 x - sin^4 x= (cos^2 x - sin^2 x)`

Replacing `cos^2 x - sin^2 x` for `cos^4 x - sin^4 x` yields:

`E = (sin^2 x + cos^2 x - sin^2 x)/(cos^2 x)`

Reducing duplicate members to numerator yields:

`E = (cos^2 x)/(cos^2 x)`

Reducing duplicate factors yields:

`E = 1`

Hence, testing if the given expression depends on `x` , yields that  does not contain the variable `x` since `E = 1` is constant.

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