1. ## Trigonometric Differentiation

Given that x = 4sin(2y+6), find (dy/dx) in terms of x.

i have managed to get:

(dy/dx) = 1/8cos(2y+6)

where do i go from here?

2. Originally Posted by Stylis10
Given that x = 4sin(2y+6), find (dy/dx) in terms of x.

i have managed to get:

(dy/dx) = 1/8cos(2y+6)

where do i go from here?
$\cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}$

I dont know what to do with the sign of cos though. It can be both plus and minus...

3. Originally Posted by Isomorphism
$\cos(2y+6) = \pm \sqrt{1 - \sin^2 (2y+6)} = \pm \sqrt{1 - \left(\frac{x}4\right)^2}$

I dont know what to do with the sign of cos though. It can be both plus and minus...
ok i under stand the first step of that, but can u please explain to me how you managed to get (x/4)^2 from sin^2(2y+6)

4. ok i understand how you got that, thanxs

5. Hello, Stylis10!

Given: . $x \:= \:4\sin(2y+6)$, .find $\frac{dy}{dx}$ in terms of $x.$

i have managed to get: . $\frac{dy}{dx} \:= \:\frac{1}{8\cos(2y+6)}$ . . . . Good!

where do i go from here?

We are given: . $4\sin(2y+6) \:=\:x\quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4} \:=\:\frac{opp}{hyp}$

So $(2y+6)$ is an angle in a right triangle with: . $opp = x,\;hyp = 4$
. . From Pythagorus, we have: . $adj = \sqrt{16-x^2}$
Hence: . $\cos(2y+6) \:=\:\frac{\sqrt{16-x^2}}{4}$

Substitute into your equation: . $\frac{dy}{dx} \;=\;\frac{1}{8\!\cdot\!\frac{\sqrt{16-x^2}}{4}}$

Therefore: . $\boxed{\frac{dy}{dx}\;=\;\frac{1}{2\sqrt{16-x^2}}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the fun of it, I did it another way.
. . (I've got to get a life!)

We have: . $4\sin(2y+6) \;=\;x \quad\Rightarrow\quad \sin(2y+6) \:=\:\frac{x}{4}$

. . . $2y + 6 \;=\;\sin^{-1}\left(\frac{x}{4}\right) \quad\Rightarrow\quad y \;=\;\frac{1}{2}\sin^{-1}\left(\frac{x}{4}\right) - 3$

Then: . $\frac{dy}{dx} \;\;=\;\;\frac{1}{2}\cdot\frac{\frac{1}{4}}{\sqrt{ 1-\left(\frac{x}{4}\right)^2}} \;\;=\;\; \frac{1}{8}\!\cdot\!\frac{1}{\sqrt{1-\frac{x^2}{16}}}$

. . . . . $\frac{dy}{dx} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\sqrt{\frac{ 16-x^2}{16}}} \;\;=\;\;\frac{1}{8}\!\cdot\!\frac{1}{\frac{\sqrt{ 16-x^2}}{4}}$

Therefore: . $\frac{dy}{dx} \;=\;\frac{1}{2\sqrt{16-x^2}}\quad\hdots$ ta-DAA!