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I am a little weak in trigonometry and was wondering if I can combine $(3w_1t)$ and $(w_1t)$ in the following function:

$$ u(t) = 20 + 64.64 \sin(w_1 t)+5\sqrt2 \sin(3w_1 t)+25\sqrt2 \cos(w_1t)+5\sqrt2 \cos(3w_1t) $$

Ideally, I'd like to get this equation in the form $$ x_n(t)=a_n \cos(nw_1t)+b_n \sin(w_1t) $$

Thank you for any insight in advance!

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    Why $x_n$ and not $x$ ? And why $nw_1$ then $w_1$ ? I suspect typos. – 2017-02-09

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I will generalize your problem a bit. We desire to simplify $$f(x)=a\sin(x)+b\sin(3x)+c\cos(x)+d\cos(3x)$$
The key to solving this comes down to noting that $$a\sin(x)+c\cos(x)=\sqrt{a^2+c^2}\sin \left(x+\arctan \left(\frac{a}{c}\right)\right)$$
If you want a proof, click on this link. Suffice to say, we set up a system of linear equalities using the "Sum of Sines" trig identity. By replacing $x$ by $3x$ we also get $$b\sin(3x)+d\cos(3x)=\sqrt{b^2+d^2}\sin \left(3x+\arctan \left(\frac{b}{d}\right)\right)$$
We thus find that $$f(x)=\sqrt{a^2+c^2}\sin \left(x+\arctan \left(\frac{a}{c}\right)\right)+\sqrt{b^2+d^2}\sin \left(3x+\arctan \left(\frac{b}{d}\right)\right)$$
You can now plug in your values to get an answer. If you really wanted to you could even turn the second sine function into a cosine by shifting the period and combine everything into one sine function, but it would probably get a lot uglier.

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    Thank you for the great explanation and links to the proof. This will definitely help! – 2017-02-09
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    @brazc0re no problem! BTW, if my solution helped you solve the problem you can mark the checkmark to accept the answer - it's generally good practice to accept the post that answered your question best, as it takes your post out if the "Unanswered Queue" – 2017-02-09
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No, you won't be able to combine terms with different periods ($\omega_1$ and $3\omega_1$).

You can, if you wish, obtain the function in the form $$ u(t) = 20 + \sqrt{64.64^2+1250} \sin\left(\omega_1t + \tan^{-1}(\frac{64.64}{25\sqrt{2}} \right) +10\sin\left(3\omega_1t +\frac{\pi}{4} \right) $$