What is the Transpose of Y = Sin(x) + Cos(x)?

[ZachGriffin]

Hi Guys, Simple question; I'm trying to work out the transpose of Y = Sin(x) + Cos(x) to make x the subject. I thought it would be x = arccos(arcsin(y)) / 2 however I don't think that's right. Is there another theorem I'm missing?

 

[Irrational]

what's y^2?

 

[ZachGriffin]

that is meant to be divided by 2

 

[Irrational]

y = sin(x) + cos(x)

\Rightarrow y^2 = (sin(x) + cos(x))^2

\Rightarrow y^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x)

\Rightarrow y^2 = \left(sin^2(x) + cos^2(x)\right) + 2sin(x)cos(x)

\Rightarrow y^2 = 1 + 2sin(x)cos(x)

\Rightarrow y^2 - 1 = 2sin(x)cos(x)

\Rightarrow y^2 - 1 = sin(2x)

\Rightarrow 2x = \sin^{-1 }(y^2 - 1)

\Rightarrow x = \frac { \sin^{-1 } (y^2 - 1) } {2}

 

[ZachGriffin]

Thanks very much for that. Anyone looking for the rules for this, I've found them on http://math2.org/math/trig/identities.htm

sin(2x) = 2 sin x cos x

sin^2(x) = 1/2 - 1/2 cos(2x)

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

 

[Irrational]

sin^2(x) + cos^2(x) = 1 should be one of the first identities you learn.

 

[Mute]

You can put it into another form by noting:

\sin x + \cos x = \sqrt{2}\left[\sin x \left(\frac{1}{\sqrt{2}}\right) + \cos x \left(\frac{1}{\sqrt{2}} \right)\right] = \sqrt{2}\left[\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right] = \sqrt{2}\sin \left(x+\frac{\pi}{4}\right).

So, y is also equal to this expression and you can solve for x. Since you don't seem to be familiar with many trig identities, I used the sine "angle sum formula", \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b.

 

[ZachGriffin]

Suppose I was to change the equation to Y = a sin(x) + b cos(x). How would I transpose it to make x the subject? I've searched through identities on that site but can't find one that relates to having 2 different coefficients a and b.

 

[Hurkyl]

Suppose I was to change the equation to Y = a sin(x) + b cos(x). How would I transpose it to make x the subject? I've searched through identities on that site but can't find one that relates to having 2 different coefficients a and b.

Think hard about how Mute's example works...

 

[ZachGriffin]

A bit more searching I've come across this:

a sin x + b cos x = R sin (x + alpha) which is an R Formulae with R = Sqrt(a^2 + b^2) and alpha = atan(b / a). If I use that I should be able to solve for x.

Having a look at Mute's post, what should I be looking for? It's been a few years since I left school so most of this is going back to that. I'd rather know how this thing works than just the answer so I'll keep looking.

 

[Hurkyl]

Having a look at Mute's post, what should I be looking for?

To derive the formula you just found by searching -- his calculation is the derivation of that formula, just in a special case.