# Period of f(t) = cos(3\pi t) + \frac{1}{2}sin(4\pi t) | T Calculation

• ChickenChakuro
In summary, the function h(x) = sin\alpha\,x, g(x) = cos\alpha\,x has a period of T=\frac{2\,\pi}{|\alpha|}.
ChickenChakuro

## Homework Statement

What is the period T of $$f(t) = cos(3\pi t) + \frac{1}{2}sin(4\pi t)$$

So, I think I have to find the LCD of 3pi and 4pi, which is 12pi. I don't think this is even close to correct though!

well here is the idea on how to show whether a function is periodic or not, but i cannot bother to look to your function, i am just too tired.
If a function is periodic then f(T+t)=f(t), where T is the period of that function. so you need to try to show that

cos[3*pi*(T+t)]+1/2sin[4*pi(T+t)]=cos(3pi t)+1/2 sin(4 pi t),
see if you can get fomething from this!

While I know this is the definition of a period, I'm not really getting it -- any other hints? Thanks.

try to use the cos(x+y)=cosxcosy-sinxsiny, and also the double angle formula for sin and cos, and see if you can cancle something out on both sides, and come up with something like this one one side

cos(..T...)sin(..T...)=0, so from here now you can let either cos(.T...)=0 or sin(..T.)=0
and determine for what value this is true, but you must have a T somewhere either in the sine or cosine or both of them. All you need to do to get there is to apply some trig identities.

Or maybe it is better to use these formulas

sinx-+siny=...
and

cosx-+cosy=...
Try both.

You know that both sine and cosine have period $2\pi$. In your specific function, $\cos(3\pi t)$ and $\sin(4\pi t)$ have periods less than $2\pi$. What are they? If you know that, then find the least common multiple of the two. Use that to find the period of the whole function.

So, the periods of the cos and sin are 3/2 and 2, respectively? I don't know if this makes any sense...

Last edited:
They should be less than $2\pi$ because the coefficient in front of t is greater than 1.

Oh, right, sorry, so the periods of cos and sin are 2/3 and 1/2, respectively, so the LCM of the two is 3?

The LCM of 2/3 and 1/2 is 2.

I think it is 3. You therefore know that if your cosine portion and your sine portion started off at a certain value each, both will be back at that certain value after t changes by 3. I don't think that constitutes a proof that 3 is the smallest number that would give this function this property, i.e. the period, but it is a good step forward at least.

I think it is 3. You therefore know that if your cosine portion and your sine portion started off at a certain value each, both will be back at that certain value after t changes by 3.

No! If $t$ changes by 3 then you have

$$f(t+3)=\cos3\,\pi\,(t+3) + \frac{1}{2}\,\sin4\,\pi\, (t+3)=\cos(3\,\pi\,t+9\,\pi)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)=-\cos(3\,\pi\,t)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)\neq f(t)$$

so the period is not 3.

In order to find the period of the above function you have to work like this.
The period of the functions $h(x)=\sin\alpha\,x,\,g(x)=\cos\alpha\,x$ is $T=\frac{2\,\pi}{|\alpha|}$. Of course every multiple of $T$ is a period too, in the sense that the functions repeat their values after $n\,T,\,n\in \mathbb{N^*}$. For the function $f$ the period of cosine is $T_1=\frac{2}{3}\quad\text{or}\quad\tau_1=\,n\frac{2}{3}$ and for the sine $T_2=\frac{1}{2}\quad\text{or}\quad\tau_2=m\frac{1}{2}$. Now in order for the two functions to have the same period, their periods must be equal, thus

$$n\frac{2}{3}=m\frac{1}{2}\Rightarrow n=\frac{3}{4}\,m$$

Thus the first integer value of $m$ which makes $n$ also an integer is $m=4\Rightarrow n=3$ and the period of $f$ is $T=2$.

Oh, that's right. I stand corrected. For some reason, when adding in my head, I managed to get 3 in my head from 2/3, but that's clearly wrong now. Don't know what I was thinking, so thanks for the catch.

## 1. What is the period of the function f(t) = cos(3πt) + 1/2sin(4πt)?

The period of a function is the length of the interval in which the function repeats itself. In this case, the function f(t) will repeat itself every π seconds. Therefore, the period of this function is π.

## 2. How do you calculate the period of a trigonometric function?

The period of a trigonometric function can be calculated by finding the smallest value of t for which the function repeats itself. This can be done by finding the smallest positive value of t for which the cosine or sine function has a value of 1. In this case, the cosine function has a period of 2π and the sine function has a period of π, so the period of the given function is the least common multiple of these two values, which is π.

## 3. Can the period of a function be negative?

No, the period of a function cannot be negative. The period of a function is a measurement of time or distance, and these values cannot be negative. However, a function can have a negative period if it is shifted on the x-axis.

## 4. How does the coefficient of the sine or cosine term affect the period of a function?

The coefficient of the sine or cosine term affects the frequency of the function, which is the number of cycles the function completes in one unit of time. The larger the coefficient, the higher the frequency and the shorter the period will be. In this case, the coefficient of the sine term is 1/2, which means the frequency is half of the cosine term, resulting in a longer period of π.

## 5. Can the period of a function change?

Yes, the period of a function can change if there is a coefficient or variable that affects the frequency of the function. In this case, if the coefficients of the sine and cosine terms were both changed, the period of the function would also change. Additionally, if the function is shifted on the x-axis, the period will also change.

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