Answer this question. Express the roots of the quartic equation for t in (b) in the form tan (n兀/16), giving the values of n. b) Given that tan (x/2)=t, express tan x and tan 2x in terms of t. Hence show that the equation tan 2x=1 may be expressed as t^4+4t^3-6t^2-4t+1.
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You should write `tan x` as `tan 2(x/2)` such that:
`tan x = tan(x/2 + x/2)`
Using the formula `tan(a+b) = (tan a + tan b)/(1 - tan a*tan b)` yields:
`tan(x/2 + x/2) = (tan (x/2) + tan (x/2))/(1 - tan^2 (x/2))`
`tan x = (2 tan(x/2))/(1 - tan^2 (x/2))`
Since the problem provides the information that `tan(x/2) = t` yields:
`tan x = (2t)/(1 - t^2)`
You need to express `tan 2x` such that:
`tan 2x = tan(x + x)`
`tan(x + x) = (tan x + tan x)/(1 - tan^2 x)`
`tan(x + x) = (2tan x)/(1 - tan^2 x)`
`tan 2x = (2*(2t)/(1 - t^2))/(1 - ((2t)/(1 - t^2))^2)`
`tan 2x = ((4t)/(1 - t^2))/(1 - (4t^2)/(1 - t^2)^2)`
`tan 2x = ((4t)/(1 - t^2))/(((1 - t^2)^2 - 4t^2)/(1 - t^2)^2)`
`tan 2x = (4t)(1 - t^2)/((1 - t^2 - 2t)(1 - t^2 + 2t))`
Checking if the equation `tan 2x = 1` may be expressed as `t^4+4t^3-6t^2-4t+1` yields:
`(4t)(1 - t^2)/((1 - t^2 - 2t)(1 - t^2 + 2t)) = 1`
`4t - 4t^3 = (1 - t^2)^2 - 4t^2`
`4t - 4t^3 = 1 - 2t^2 + t^4 - 4t^2`
`4t - 4t^3 = t^4 - 6t^2 + 1`
Moving all terms to one side yields:
`t^4 + 4t^3- 6t^2 - 4t+ 1 = 0`
Comparing the given equation in t to the equation that results substituting `(4t)(1 - t^2)/((1 - t^2 - 2t)(1 - t^2 + 2t))` for `tan 2x` yields that `t^4 + 4t^3 - 6t^2 - 4t + 1 = t^4 + 4t^3 - 6t^2 - 4t + 1` , hence, the alternative form of the equation `tan 2x = 1` is `t^4 + 4t^3 - 6t^2 - 4t + 1 = 0` .
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