PLEASE HELP ME !!!!! How do I turn this cos function into a sin function : 19 cos(0.104t)21

Question

tkhunny · Answer

You're going to have to enter it better than that.

sin(a) = cos(pi/2 - a)

anonymous · Answer

sorry 19 cos(0.104 t) + 21

anonymous · Answer

So will it be 19 sin(0.104(t-pi)) +21?

tkhunny · Answer

$\cos(0.104t) = \sin\left(\dfrac{\pi}{2} - 0.104t\right)$

anonymous · Answer

so when you graph both of them they should look the same right ?

tkhunny · Answer

And indeed they do!

anonymous · Answer

they do not look a like

anonymous · Answer

l don't understand

tkhunny · Answer

Then you did not graph them both correctly.

This is no good:  sin(0.104(t-pi))

This is right: sin(pi/2 - 0.104t)

anonymous · Answer

Okay but where did you find pi/2 ?

anonymous · Answer

like is there a mathematical way to find that shift ?

tkhunny · Answer

Way back in your trigonometry class, you were forced to memorize and manipulate identity after identity after identity after identity.  Are you remembering this with fondness?  This is just one of those many identities.

$\sin(x) = \cos(\pi/2 - x)$

It's just an identity.  You can use $\sin^{2}(x) + \cos^{2}(x) = 1$ if you like.  I just thought I'd pick an easier one and save some trouble.

anonymous · Answer

okay so I insert the cos and I will get that

tkhunny · Answer

Pretty hard to confirm via rumor.  Show me what you are planning.

anonymous · Answer

sin^2 +cos(0.104) ^2 +1

tkhunny · Answer

??  That makes no sense at all.

You have $\sin^{2}(x) + \cos^{2}(x) = 1$

This is an identity.  As long as those arguments are the same, the sum of the two functions will be 1.

$\sin^{2}(2) + \cos^{2}(2) = 1$

$\sin^{2}(\sqrt{2}) + \cos^{2}(\sqrt{2}) = 1$

$\sin^{2}(Fred) + \cos^{2}(Fred) = 1$

$\sin^{2}(0.104t) + \cos^{2}(0.104t) = 1$

If you have $\cos(0.104t)$, you must solve the identity for that value or function.

$\cos^{2}(0.104t) = 1 - \sin^{2}(0.104t)$

$\cos(0.104t) = \sqrt{1 - \sin^{2}(0.104t)}$ -- Although, you have to be a little careful with that, since we just messed with the Domain.

I highly recommend the original substitution.

anonymous · Answer

Oh I see thank you !

tkhunny · Answer

??  That makes no sense at all.

You have $\sin^{2}(x) + \cos^{2}(x) = 1$

This is an identity.  As long as those arguments are the same, the sum of the two functions will be 1.

$\sin^{2}(2) + \cos^{2}(2) = 1$

$\sin^{2}(\sqrt{2}) + \cos^{2}(\sqrt{2}) = 1$

$\sin^{2}(Fred) + \cos^{2}(Fred) = 1$

$\sin^{2}(0.104t) + \cos^{2}(0.104t) = 1$

If you have $\cos(0.104t)$, you must solve the identity for that value or function.

$\cos^{2}(0.104t) = 1 - \sin^{2}(0.104t)$

$\cos(0.104t) = \sqrt{1 - \sin^{2}(0.104t)}$ -- Although, you have to be a little careful with that, since we just messed with the Domain.

I highly recommend the original substitution.