# Thread: Calculating derivitive from trig equation

1. ## Calculating derivitive from trig equation

Hi, i need to calculate the derivitive from the following formular:

y = 3sin (PIE * t / 12) : t is a variable and PIE is 3.14 etc

I think its: y = PIEcos (PIE * t / 12)

Or would it be: y = 3PIEcos (PIE * t / 12) ??

All help is appreciated.

2. Hello,

Let a be pi/12 (no e at the end of pi...).

Your function is now 3 sin(a*t)
Its derivative if 3*a*cos(a*t)

That is to say 3*pi/12 *cos(pi*t/12)=pi/4*cos(pi*t/12)

3. ## Ok here you go

if $f(x)=3sin\bigg(\frac{\pi{t}}{12}\bigg)$...then $f'(x)=3cos\bigg(\frac{\pi{t}}{12}\bigg)*\frac{\pi} {12}=\frac{\pi}{4}cos\bigg(\frac{\pi{t}}{12}\bigg)$

4. Thanks for your replies. The next step asks, "calculate the rate of change of the depth of the water when t = 19."

Would the answer to this be 0.203?? (to 3dp)

would be $f'(19)=\frac{\pi{cos\bigg(\frac{19\pi}{12}\bigg)}} {4}=\frac{(\sqrt{3}-1)\pi{\sqrt{2}}}{16}=.0677$

6. Awesome, thanks. But if you follow your equation through, i get the answer as 0.575?? Also, where did the square root of 3 come from?

7. I think that you can have the exact value of cos(19pi/12) by saying that 19pi/12=3pi/2+pi/12

After that, i don't really know, you can use some methods you study in high school, but it's kinda too far for me

8. ## Wow

I am sorry I am the typo king today it should be $\frac{(\sqrt{3}-1)\pi{\sqrt{2}}}{16}$...and this comes from using the basic unit circle

9. I'd like to see your proof that cos(19pi/12) equals to what you said
Not that i think it's false, but it's a method i've forgotten...

10. Sweet, thanks for all your help. Are you able to just quickly explain where you get the square root bits from? It is confusing me a bit.

11. ## Its kinda complicated

...do you really want me to go through it?

12. Oh yeah ^^

If you could give the formula so quickly, it may not be so difficult to tell us how to do ?

13. ## hah

ok so persistent $cos\bigg(\frac{19\pi}{12}\bigg)=cos\bigg(\frac{3\p i}{2}+\frac{\pi}{12}\bigg)=cos\bigg(\frac{3\pi}{2} \bigg)cos\bigg(\frac{\pi}{12}\bigg)-sin\bigg(\frac{3\pi}{2}\bigg)sin\bigg(\frac{\pi}{1 2}\bigg)$ $=0-\bigg(\frac{(\sqrt{3}-1)\sqrt{2})}{4}\bigg)(-1)$..thes being commonly known ones...and now multiplying that by $\frac{\pi}{4}$ gives you what you want..now before you ask $sin\bigg(\frac{\pi}{12}\bigg)$ is just one you have to know its just one of the exact values on the unit circle

14. is just one you have to know
Really ?
The commonly learnt values are pi/6, pi/4, pi/3, pi/2 (and 0, pi etc...), but it's not so easy (let's say 3 lines of calculus) to retrieve it with pi/12=1/2*pi/6

15. ## if you are trying to be sardonic

if you did rewrite it as $sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg)$ you would nowhere to go except use another trig identity that $sin(2x)=(-1)^{\frac{x}{2\pi}}\sqrt{\frac{1-cos(x)}{2}}$..then you'd have to go that $cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}$...then go that the $sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg)=$ is equal to $\frac{\pi}{6}$ imputed into that forumla...for me its easier to remember what the $sin\bigg(\frac{\pi}{12}\bigg)$ is

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