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.
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.
ok so persistent $\displaystyle 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)$$\displaystyle =0-\bigg(\frac{(\sqrt{3}-1)\sqrt{2})}{4}\bigg)(-1)$..thes being commonly known ones...and now multiplying that by $\displaystyle \frac{\pi}{4}$ gives you what you want..now before you ask $\displaystyle sin\bigg(\frac{\pi}{12}\bigg)$ is just one you have to know its just one of the exact values on the unit circle
if you did rewrite it as $\displaystyle sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg)$ you would nowhere to go except use another trig identity that $\displaystyle sin(2x)=(-1)^{\frac{x}{2\pi}}\sqrt{\frac{1-cos(x)}{2}}$..then you'd have to go that $\displaystyle cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}$...then go that the $\displaystyle sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg)=$ is equal to$\displaystyle \frac{\pi}{6}$ imputed into that forumla...for me its easier to remember what the $\displaystyle sin\bigg(\frac{\pi}{12}\bigg)$ is