Question

Given f(x)=cosxf(x)=cosx, which function below doubles the amplitude and has a period of 3π3π?g(x)=3cos2xg of x is equal to 3 cosine 2 xg(x)=12cos2xg of x is equal to 1 half cosine 2 xg(x)=2cos2x3g of x is equal to 2 cosine 2 x over 3g(x)=3cos3x2g of x is equal to 3 cosine 3 x over 2

Answer 1

The derivation that correctly uses the cosine sum identity to prove the cosine double angle identity is A. A 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = cosine (x) cosine (x) minus sine (x) sine (x), = cosine squared (x) minus sine squared (x)

Step-by-step explanation:

It should be noted that the cosine difference identity is found by simplifying the equation by first squaring both sides.

Therefore, the derivation that correctly uses the cosine sum identity to prove the cosine double angle identity is that a 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = cosine (x) cosine (x) minus sine (x) sine (x), = cosine squared (x) minus sine squared (x).

In conclusion, the correct option is A.

Answer 2

`g(x)=2\cos (2x)/(3)`

Step-by-step explanation:

A cosine function is generally given as;

`\begin{gathered} y=a\cos (b) \\ \text{where Amplitude }=|a| \\ \text{ Period }=(2\pi)/(|b|) \end{gathered}`

Given the below function;

`f(x)=\cos x`

If we compare both functions, we'll see that a = 1 and b = 1.

If we need another function with double the amplitude, then the value of a in that function will be (a = 2 x 1 = 2).

If we're to have another function g(x), with a period of 3 pi, let's go ahead and determine the value of b in the second function;

`\begin{gathered} (2\pi)/(b)=3\pi \\ 3\pi\cdot b=2\pi \\ b=(2\pi)/(3\pi) \\ b=(2)/(3) \end{gathered}`

Since we now have that for the second function g(x), a = 2 and b = 2/3, therefore g(x) can be written as below;

`g(x)=2\cos (2x)/(3)`