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I have a link to a paper on a solution below

http://math.berkeley.edu/~zworski/128/psol07.pdf

This is related to my other question on the same problem.

For problem 7, the author achieves a second error bound by using

$$f^4(\xi) = | -119e^{2\xi}sin(3\xi) - 120e^{2\xi}cos(3\xi)|$$

and concludes

$$f^4(\xi) = | -119e^{2\xi}sin(3\xi) - 120e^{2\xi}cos(3\xi)| \le e^4\sqrt{119^2 + 120^2} \le 120\sqrt{2}e^4$$

What is the author doing here to get

$$e^4\sqrt{119^2 + 120^2}$$?

If someone could explain to me what is going on you'd save me a lot of time. I'm not sure what properties of trigonometric functions he's capitalizing on.

thanks!

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  • $\begingroup$ I'm not exactly sure on that step, but the last part can be achieved by using $|sin(x)+cos(x)|\leq \sqrt{2}$ $\endgroup$
    – Thomas
    Nov 23, 2014 at 21:07

1 Answer 1

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Use that $a\sin x + b\cos x = \sqrt{a^2 + b^2}\sin\left( x + \text{atan2}(b,a) \right)$.

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  • $\begingroup$ What is the name of this identity? I have never heard of it before. I looked up the proof, very interesting. Once he established this bound I'm assuming he used $sin(x) \le 1$ for all reals, which just left him with $\sqrt{a^2 + b^2}$ $\endgroup$
    – John
    Nov 23, 2014 at 23:11
  • $\begingroup$ Correct. I would call it linear combination of sines or phasor addition. $\endgroup$
    – slo
    Nov 24, 2014 at 6:59

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