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Linear trigonometric equation

Posted September 24th, 2008 by Structure

We solve the equation

$a\sin x+b\cos x+c=0$

where $a,b,c \in R$ and $a^2+b^2>0$ .
Then we can write

$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x+\frac{c}{\sqrt{a^2+b^2}})=0$

There is a unique $t\in[0,2\pi)$ such as

$\cos t=\frac{a}{\sqrt{a^2+b^2}}$

and

$\sin t=\frac{b}{\sqrt{a^2+b^2}}$

.
so we can write

$\sqrt{a^2+b^2}(\sin x\cos t+\cos x\sin t+\frac{c}{\sqrt{a^2+b^2}})=0$

$\sqrt{a^2+b^2}(\sin (x+t)+\frac{c}{\sqrt{a^2+b^2}})=0$

Now, if

$|\frac{c}{\sqrt{a^2+b^2}}|\leq 1$

we have a countable set of solutions

Posted July 24th, 2008 by Structure

Let $a,b\in N$ such as $a^2-b=c^2$ for some $c\in N$ Then

$\sqrt{a\pm \sqrt b}=\sqrt{\frac{a+c}{2}}\pm \sqrt{\frac{a-c}{2}}$

It is easy to verify as

$a\pm \sqrt b=\frac{a+c}{2}}+\frac{a-c}{2}\pm 2\sqrt{\frac{a^2-c^2}{4}}=a\pm \sqrt b$

An example :for a=2 and b=3 we have c=1 so

$\sqrt{2\pm \sqrt 3}=\sqrt{\frac{2+1}{2}}\pm \sqrt{\frac{2-1}{2}}=\sqrt{\frac{3}{2}}\pm \sqrt{\frac{1}{2}}$

a=6 and b=11 we have $c^2=36-11=25$ so $c=5$

$\sqrt{6\pm \sqrt 11}=\sqrt{\frac{6+5}{2}}\pm \sqrt{\frac{6-5}{2}}=\sqrt{\frac{11}{2}}\pm \sqrt{\frac{1}{2}}$

Posted July 5th, 2008 by Structure