Write a function whose parent is y=sin x and whose graph has amplitude 3/4 and period pi

Give a general solution to cos^2 theta=3/4

What are the period, amplitude, and shifts of the graph y= 3 cos (2x-pi) +5 from y=cos x

Thank you!!

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- Jul 6th 2013, 07:59 AMawesomest47Functions, Statistics, Trigonometry Questions?
Write a function whose parent is y=sin x and whose graph has amplitude 3/4 and period pi

Give a general solution to cos^2 theta=3/4

What are the period, amplitude, and shifts of the graph y= 3 cos (2x-pi) +5 from y=cos x

Thank you!! - Jul 6th 2013, 09:21 AMHallsofIvyRe: Functions, Statistics, Trigonometry Questions?
What exactly is your difficulty? Are you saying that you have never taken trigonometry and don't know the meanings of the words? Or do you just not

**want**to try anything? - Jul 6th 2013, 01:43 PMawesomest47Re: Functions, Statistics, Trigonometry Questions?
Excuse me for actually thinking the math help forum was intended for finding assistance with my homework assignments I struggle with. Rather than insulting the fact that I have a question and accusing me of being a lazy, perhaps you could actually guide me in finding the correct answer. Thank you for the

*kind*assistance you have provided to be. - Jul 6th 2013, 06:58 PMProve ItRe: Functions, Statistics, Trigonometry Questions?
- Jul 8th 2013, 08:18 AMawesomest47Re: Functions, Statistics, Trigonometry Questions?
Oh, thank you. Sorry this is my first post! My teacher helped me with those problems but I have another question and was wondering if someone could help me with it. There is a cosine graph with three points on it (0,30) (6,83) (12,30) and I need to write a cosine function. So far, I have y=56.5 cos (12/2pix- ?) but I have no idea how to find the phase shift to complete this function based on the given graph. There is no feature like this on a Ti-Nspire CAS.

Also, I need to find the inverse sine of 1/2 and the inverse tangent the square root of 3. So I am pretty sure this relates back to ratios in a 30 60 90 triangle but I don't know what to do with these numbers to write these functions in terms of pi. Do I need to multiply by each ratio but how do I know what side is which?

i hope I asked my question correctly this time and somebody can give me some help! Thank you!! - Jul 8th 2013, 08:23 AMProve ItRe: Functions, Statistics, Trigonometry Questions?
Your general cosine function is of the form $\displaystyle \displaystyle y = a\cos{\left[ b \left( x - c \right) \right] } + d $. The standard method would be to substitute four points to give four equations to solve in your CAS, but you only have been given three. So you must have more information.

I see that you have the points (0, 30) and (12, 30). Is this just one cycle of your graph or have there been more cycles? - Jul 9th 2013, 12:17 PMawesomest47Re: Functions, Statistics, Trigonometry Questions?
There is only one cycle. When I asked my teacher what I should do, all he said was "Shouldn't need a calc to find the equation, use what you know about amplitude, period and stuff to find it." I have attached a photo of the entire question (7.) Hope it helps- thank you!

Attachment 28770 - Jul 9th 2013, 12:19 PMawesomest47Re: Functions, Statistics, Trigonometry Questions?
Also, can somebody help me understand using a 30 60 90 triangle to find inverse trigonometric functions (e.x. inverse sine of 1/2 and inverse tangent of the square root of 3.) Thank you!!

- Jul 10th 2013, 09:45 AMawesomest47Re: Functions, Statistics, Trigonometry Questions?
Anybody help me with this cosine graph?

- Jul 10th 2013, 05:09 PMtopsquarkRe: Functions, Statistics, Trigonometry Questions?
Please do not bump your posts. (See the forum rules below.)

It's a temperature graph so it is cyclic. By inspection you can see that if you have the point (6, 83) you will also have the point (18, 83). Do you see why?

Also, tell us if the form $\displaystyle y = a~cos(b[x - c]) + d$ or the more "Physics-y" one $\displaystyle y = A~cos( \lambda x - \phi ) + A_0$. The first is a "curve-fitting" task, whereas the second is more of a "physical" one.

-Dan