Question

Help, please?

If `cos(2pi/5)=x/2` Show that `x^2+x-1=0`

Answer 1

Please see below.

Explanation:

We know that,

`color(red)(sin18^0=(sqrt5-1)/4`

Here,

`cos((2pi)/5)=x/2`

We have,

`(2pi)/5=360^0/5=72^0`

So,

`cos72^0=x/2`

`=>cos(90^0-18^0)=x/2`....`.But, cos(pi/2-theta)=sintheta`

`=>sin18^0=x/2`
`=>(sqrt5-1)/4=x/2`

`(sqrt5-1)/2=x`

`=>sqrt5-1=2x`

`=>sqrt5=2x+1`

Squaring both sides

`(sqrt5)^2=(2x+1)^2`

`=>5=4x^2+4x+1`

`=>4x^2+4x-4=0`

Dividing both sides by 4

`=>x^2+x-1=0`

Answer 2

Please go through a Proof in the Explanation.

Explanation:

Given that, `cos(2/5pi)=x/2 :. x=2cos(2/5pi)=2cos72^@`.

Hence. `x^2+x=4cos^2 72^@+2cos72^@`,

`=2{1+cos(2xx72^@)}+2cos72^@`,

`=2(1+cos144^@+cos72^@)`,

`=2{1+cos(180^@-36^@)+cos72^@}`,

`=2{1-cos36^@+cos72^@}`,

`=2{1+(cos72^@-cos36^@)}`,

`=2{1-2sin((72^@+36^@)/2)sin((72^@-36^@)/2)}`,

`=2{1-2sin54^@sin18^@}`,

`=2{1-(2sin54^@sin18^@cos18^@)/cos18^@}`,

`=2[1-{sin54^@(2sin18^@cos18^@)/cos18^@}]`,

`=2{1-2/2xx(sin54^@sin36^@)/cos18^@}`,

`=2{1-(2sin54^@sin36^@)/(2cos18^@)}`,

`=2[1+{cos(54^@+36^@)-cos(54^@-36^@)}/(2cos18^@]`,

`=2{1+(cos90^@-cos18^@)/(2cos18^@)}`,

`=2{1+(0-cos18^@)/(2cos18^2)}`,

`=2{1-1/2}`.

`rArr x^2+x=1, or, x^2+x-1=0`, as desired!

N.B.:

The Problem can easily be solved using the value of

`x=2cos72^@=2sin18^@=2*1/4(sqrt5-1)=(sqrt5-1)/2`.

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