# Change the subject

## Homework Statement

make $$\theta$$the subject
$$y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2$$

## The Attempt at a Solution

not sure how to get start :(
should i change tan to sin/cos first?

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tiny-tim
Homework Helper
Welcome to PF!

## Homework Statement

make $$\theta$$the subject
$$y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2$$

not sure how to get start :(
should i change tan to sin/cos first?
Hi suppy123! Welcome to PF!

You ultimately want to find an equation beginning "θ =".

But the last-step-but-one will be an equation beginning "cosθ =" or "tanθ =".

So the technique you need is to change the given equation so that it either has only cosθ (and no tanθ) or only tanθ (and no cosθ).

In other words: either rewrite tanθ in terms of cosθ, or cosθ in terms of tanθ (whichever seems easier).

hi,
ok,
$$y = x(\frac{\sin{\theta}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{1}{\cos^2{\theta}})$$

$$y = x(\frac{\cos{\theta+\frac{\pi}{2}}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{2}{1+\cos{2{\theta}}})$$

is this correct? :S

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tiny-tim
Homework Helper
Nooo …

You need the whole of "cosθ", not just the "cos" part.

An equation with both cosθ and cos(π/2 - θ) won't do.

Hint: tanθ is easier. How can you express 1/cos²θ in terms of tanθ?

yup, it's $$\tan^2{\theta}+1$$
$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)$$
$$\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}$$
$$\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}$$

so
$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{2\tan{\theta}}{\tan{2\theta}}+2)$$

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tiny-tim
Homework Helper
yup, it's $$\tan^2{\theta}+1$$
ok, so the original equation in terms of tanθ is … ?

is that right?

sry it should be
$$\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1$$

$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)$$

tiny-tim
Homework Helper
sry it should be
$$\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1$$

$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)$$
oh nooo … you've gone completely off the rails …

It's $$\tan^2{\theta} = \frac{-2\tan{\theta}}{\sin{2\theta}}+1$$

Your original equation 1/cos²θ = tan²θ +1 was right.

But don't start using tan2θ … having both tan2θ and tanθ is as bad as having both cosθ and cos(π/2 - θ).

You want an equation with only tanθ, tan²θ, tan³θ, and so on.

$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}+1)(\tan{\theta}-1)+2$$

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tiny-tim
Homework Helper
$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}-\sqrt{1})^2+1$$
ooo …

I've just noticed that you did produce the right equation (before you went berserk):
$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)$$
That's the equation you want!

Now that's a quadratic equation in tanθ, so use the usual -b ± √etc formula, to get an equation starting "tanθ ="

lol...sweet, so
$$y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)$$
$$0 = (-\frac{gx^2}{2vi^2})(\tan^2{\theta}) + x(\tan{\theta}) - (\frac{gx^2}{2vi^2} - y)$$
$$\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2}-y)^2}}{-2\frac{gx^2}{2vi^2-y}}$$

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oops, forgot to change the y heh

tiny-tim
Homework Helper
$$\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2})^2}}{-2\frac{gx^2}{2vi^2}}$$
erm …
What happened to poor little y ?

there you go
$$\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)^2}}{-2(\frac{gx^2}{2vi^2})}$$
;0

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tiny-tim
Homework Helper
ooh … there he is … !

hmm … that last equation doesn't look right, though.

should be right this time ;)
$$\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)}}{2(-\frac{gx^2}{2vi^2})}$$

tiny-tim