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I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain?

from

$$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$

transform to

$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t\left(\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\cos(2t)+\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}\sin(2t)\right)$$

let $\displaystyle\frac{1}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\cos\phi$ and $\displaystyle\frac{\frac{1}{2}}{\sqrt{1^2+\left(\frac{1}{2}\right)^2}}=\sin\phi$,

$$f(x)=\frac{3}{5}-\frac{3}{5}\sqrt{1^2+\left(\frac{1}{2}\right)^2}e^t(\cos\phi\cos(2t)+\sin\phi\sin(2t))$$

Armando j18eos
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Cable W
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    You have the expression $1\cdot \cos 2t+ \frac 12\cdot \sin 2t$ from which $1$ and $\frac 12$ can be read off immediately. This is where they come from. Answers will explain more about why. – Mark Bennet Aug 17 '20 at 08:24

2 Answers2

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Let’s concentrate on the important part, which is of the form $$ f(x)=a\cos x+b\sin x $$ which we want to express as $$ f(x)=A(\cos\varphi\cos x+\sin\varphi\sin x) $$ A necessary (and sufficient) condition is that $$ A\cos\varphi=a,\qquad A\sin\varphi=b $$ and therefore $a^2=A^2\cos^2\varphi$, $b^2=A^2\sin^2\varphi$. Hence $$ A^2=a^2+b^2 $$ We want $A>0$ (not necessary, but convenient), so we get $$ A=\sqrt{a^2+b^2},\quad \cos\varphi=\frac{a}{A},\quad \sin\varphi=\frac{b}{A} $$ The last two requirements can be fulfilled, because $(a/A,b/A)$ is a point on the unit circle.

egreg
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  • Thank you :) $sin^2 +cos^2 = 1$ => $a^2 + b^2 = A^2(cos^2 +cos^2 ) = A^2$

    also $\frac{b^2}{a^2}=\frac{sin^2 }{cos^2 } = tan^2 => \frac{b}{a}=tan$

     – Cable W Aug 17 '20 at 10:06
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    @CableW The last passage is good, but writing $\varphi=\arctan(b/a)$ would be incorrect, because the arctangent only returns angles in the first and fourth quadrant. If $a<0$ you need to take $\pi+\arctan(b/a)$. – egreg Aug 17 '20 at 10:18
  • Got it, thank you! – Cable W Aug 17 '20 at 10:23
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This is a way to normalize the vector $v=(a,b)=\left(1,\frac12\right)$ that is

$$|v|=\sqrt{a^2+b^2} \implies \hat v=\frac{v}{|v|}$$

has length equal to $1$ and this allows to perform the subsequent transformation for $\cos \phi$ and $\sin \phi$.

user
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  • Thank you! I guess that I still need to figure out the relationship between sinusoids and vector normalization. Go searching on Google now :) – Cable W Aug 17 '20 at 08:52
  • @CableW You are welcome! Good work. Bye – user Aug 17 '20 at 08:57