- anonymous

thanks

- chestercat

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- anonymous

You are going to want to use a double angle identity

- anonymous

Do you still remember you Double angle formulas?

- anonymous

kind of

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## More answers

- anonymous

If you don't you can quickly find them on the net with a quick search :)

- anonymous

got it then what

- anonymous

From there, it is usually trial and error to see what you get in the end, how would you start this one off ?

- anonymous

Use a double angle formula for the left side and convert the right side in terms of sin and cos

- anonymous

So straight away we can change Sin(2x) into 2SinxCosx

- anonymous

2cos(2x)/sin(2x)= 2sinxcosx

- anonymous

No just Sin(2x) = 2SinxCosx

- anonymous

The Cos(2x) is something we have to play around with, since there are three variants of this particular double angle

- anonymous

x=0.15+0.2 dot n

- anonymous

Ummm... I not entirely sure how you got that ...

- anonymous

In this question we are proving that LHS = RHS not finding a value for x

- anonymous

ow got it so what are we doing next

- anonymous

Alright, I'll show you what I got, and lets work from there

- anonymous

thanks

- anonymous

= 2Cos(2x)
Sin2(x)
= 2[1-sin^2x]
2SinxCosx

- anonymous

is that final answer

- anonymous

Not even haha, its just the first step ^^

- anonymous

lol then what

- anonymous

So from there
-> Expand inwards
2 - 2Sin^2x
2SinxCox

- anonymous

No split the fraction into:
2 - 2Sin^2x
2SinxCosx 2SinxCosx

- anonymous

so sin^x

- anonymous

Firstly we'll solve this part first:
2
2SinxCosx

- anonymous

From here, we can clearly cancel the two's out ^^
So it will become
1
SinxCosx

- anonymous

So you say to your self, how do we get tanx? out of
1
SinxCosx
well remember that sinx / cosx = tanx

- anonymous

Be careful. You can't get tan out of 1/sinxcosx

- anonymous

We have to move the Sinx into the numerator, by taking the reciprocal of that. Note that the reciprocal is different from inverse

- anonymous

Just like this mate,
(Sinx)^-1
Cos
(Tanx)^-1 = Tan^-1x
-> Apply the inverse of Tan to get Cot
Cotx

- anonymous

Sorry I meant, the reciprocal of tan to get cot ><

- anonymous

its all good thanks

- anonymous

Now we have to fine the other part, which is straight forwards stuff :)
2Sin^2x
2SinxCosx

- anonymous

So we have to get Tanx out of 2Sin^2x
2SinxCosx

- anonymous

How would you approach in doing this?

- anonymous

get rid of two 2sin and be left with xcosx?

- anonymous

if only XD
But your procedure is a bit off ><
I'll show you :)

- anonymous

lol thanks

- anonymous

So we can cancel out the two's to get
2Sin^2x = Sin^2x
2SinxCosx SinxCosx

- anonymous

Because the two's are just coefficients in front of the terms, they don't act like terms themselves :)

- anonymous

Anyways, from there
notice how Sin^2x has a power of 2
That means Sinx . Sinx = Sin^2x

- anonymous

So by working out we get
Sinx . sinx
Sinx Cosx

- anonymous

We can easily cancel the Sinx / sinx to get 1

- anonymous

And finally
1 . Sinx = Tanx
Cosx

- anonymous

And so we have proven that
2cos(2x)/sin(2x)=cot(x)-tan(x)

- anonymous

so its cosx :)

- anonymous

Ah just remember that Sinx/cosx = tanx :)

- anonymous

thanks so much!!!!!!!!!!!!!!!!!!!!!!!

- anonymous

no problem !

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