Question

If x+1/x=2cosθ,then prove that x=cosθ±isinθ

Answer 1

\[x+\frac{ 1 }{ x }=2\cos \theta\]

Answer 2

we have to prove that it equals to

Answer 3

\[x=\cos \theta \pm i \sin \theta\]

Answer 4

x^2-2xcos theta +1=0
solve this using formula

Answer 5

@hartnn i did'nt get it

Answer 6

note this:
we know x+ 1/x >= 2

hence 2cos@ >=2
cos@ >=1
only possibility with real solutions will be cos@ = 1 .

for your question,
we have on re-arranging ,
x^2 -2x cos@ + 1=0

x= [2cos@+-sqrt(4cos^2 @ -4) ]/2
= cos@ +- sqrt(-sin^2 @)
= required solution..

Answer 7

then look ^^

Answer 8

how did x+1/x=2 ?? i am getting confused ;(

Answer 9

forget that..on re-arranging, you get a quadratic eqn, just concentrate on that..

Answer 10

and thats x+ 1/x >= 2 and not =2//but still,,dont think too much about it right now..

Answer 11

u know i am not used to of these trignometric functions like sin or cos

Answer 12

thats why i find this question tuff

Answer 13

where do you get stuck?

Answer 14

from the start

Answer 15

i just dont get the logic of question

Answer 16

first you multiply by x:
\[x^2 + 1 = 2\cos \theta x\]

Answer 17

man i am exploding goku please help

Answer 18

All it is doing is asking you to solve for x

Answer 19

yes

Answer 20

you just have trigonometric terms as the coefficients of your quadratic equation

Answer 21

yeah

Answer 22

what should i do next after x^2+1=2costhetax

Answer 23

so you have the equation:
\[x^2 -2\cos \theta x +1\]
where a=1, b=-2cos(theta), c = 1

Answer 24

just subtract 2cos(theta)

Answer 25

so you have the above quadratic, and you can still use the quadratic formula to solve for the roots

Answer 26

ok wait let me do it

Answer 27

i am stuck

Answer 28

I am too learning this Mr moose u good at this well done

Answer 29

\[\frac{-(-2\cos \theta) \pm \sqrt{(-2\cos \theta)^2 -4}}{2}\]

Answer 30

can you solve this?

Answer 31

@Miyuru thank you

Answer 32

\[x=\frac{ 2\cos \theta \pm \sqrt{4\cos^2 \theta-4} }{2}\]

Answer 33

yes

Answer 34

i am stuck here

Answer 35

U are well come All the best help @koli123able to solve this i am not good at this hope u the best..

Answer 36

ok, so lets look at the discriminant:
\[4\cos^2\theta-4\]
you can factor out a four:
\[4(\cos^2\theta -1)\]

Answer 37

oh!

Answer 38

now, do you remember the Pythagorean identity?

Answer 39

sin^2 theta

Answer 40

almost!

Answer 41

be careful about that

Answer 42

-sin^2(theta)

Answer 43

do you think you can figure it out now?

Answer 44

ok wait

Answer 45

\[\sqrt{4-\sin^2 \theta }\] stuck here

Answer 46

\[\frac{2\cos(\theta) \pm \sqrt{-4\sin^2 \theta}}{2}\]

Answer 47

oh!

Answer 48

but still i am stuck how will i get its square root

Answer 49

is it the fact that the square root is imaginary?

Answer 50

Mr moose is this to be solved using the quadratic formula....can't it be solved using quadratic equation..

Answer 51

\[2 \sin \theta i\]

Answer 52

yes

Answer 53

so cancel out all of the twos and you get...

Answer 54

got got got it

Answer 55

Well done @MrMoose u are great/.

Answer 56

thank you

Answer 57

So, in general, when you are solving these types of problems, remember that cos(theta) and sin(theta) and the like can be manipulated like constants

Answer 58

Also, be sure you know your identities.

Answer 59

this is touching :)

Answer 60

:)