Question

Using a trig identity, write (equation in question) using only one cosine function

Answer 1

x(t)=5cos(4t)-2sin(4t)

Answer 2

Use the identity
\[\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\]

Answer 3

so, the alpha and betas are both 4t right? What do I do with the integers before the trig functions?

Answer 4

If both alpha and beta were 4t, you'd have
\[\cos0=\cos^24t+\sin^24t,\]
which, although true, doesn't help you. I'll try to work it out by hand for myself; I haven't done this particular problem in a while.

Answer 5

oh wow right, so I must have to split up the (4t) into 2 components somehow, to get and alpha and beta, and then find the difference between them.....

Answer 6

i found a trig identity that should help, but i don't know how to get it into one cosine function.
\[Acos(wt)+Bsin(wt)=Ccos(wt-\gamma)\]

Answer 7

where
\[C=\sqrt(A^2+B^2)\]
sin(gamma)=B/C
cos(gamma)=A/C

Answer 8

Starting with the identity, and letting k be some constant:
\[k\cos(\alpha-\beta)=k\cos\alpha\cos\beta+k\sin\alpha\sin\beta\]
You want the right side to look like the given expression for x(t), so we'll equate the right side to it:
\[k\cos\alpha\cos\beta+k\sin\alpha\sin\beta=5\cos(4t)-2\sin(4t)\]
Letting alpha = 4t,
\[k\cos\beta\cos(4t)+k\sin\beta\sin(4t)=5\cos(4t)-2\sin(4t)\]
Matching up the coefficients, you have
\[\begin{cases}k\cos\beta=5\\
k\sin\beta=-2\end{cases},\text{ or equivalently, }\begin{cases}\cos\beta=\frac{5}{k}\\
\sin\beta=\frac{-2}{k}\end{cases}\]
Draw up a reference triangle that satisfies the system:

Solving for k yields
\[k^2=5^2+(-2)^2\Rightarrow k=\sqrt{29}\]
So, you're left with
\[\begin{align*}k\cos(\alpha-\beta)&=5\cos(4t)-2\sin(4t)\\
k\cos\beta\cos(4t)+k\sin\beta\sin(4t)&=\\
\sqrt{29}\cos(4t-\beta)&=\end{align*}\]
From the triangle you know that
\[\tan\beta=\frac{-2}{5},\text{ so }\beta=\tan^{-1}\frac{-2}{5}\]
So, you get
\[x(t)=\sqrt{29}\cos\left( 4t-\tan^{-1}\left(-\frac{2}{5}\right)\right)\]

Answer 9

i got down to
\[\sqrt(29)*\cos(4t-\cos^{-1} (-2/\sqrt29)\]
but the question wants the answer in terms of only one cosine function

Answer 10

and unfortunately the same holds true for your answer :(

Answer 11

It's still technically in terms of cosine. The inverse tangent term in the result is just a number.

Answer 12

woah ok so i put in your answer any way and it worked....thanks you so much

Answer 13

tan(some number) would also be some number. tan(t) is a function of t.

Answer 14

ok I get it! thank you.....how do you give medals cuz you deserve 2 haha