Dexter Odom

2022-03-28

$-\mathrm{sin}x-\mathrm{cos}x=-\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
How does the cosine disappear and how did $\mathrm{sin}x$ turn into $\mathrm{sin}\left(x+\frac{\pi }{4}\right)$

kachnaemra

A very useful formula:
$a\mathrm{sin}x+b\mathrm{cos}x=\sqrt{{a}^{2}+{b}^{2}}\mathrm{cos}\left(x-\mathrm{arctan}\left(\frac{a}{b}\right)\right)$
$\sqrt{{a}^{2}+{b}^{2}}\mathrm{cos}\left(x\right)\mathrm{cos}\left(\frac{\mathrm{arctan}a}{b}\right)+\mathrm{sin}\left(x\right)\mathrm{sin}\left(\frac{\mathrm{arctan}a}{b}\right)$
$=\frac{b\sqrt{{a}^{2}+{b}^{2}}\mathrm{cos}\left(x\right)}{\sqrt{{a}^{2}+{b}^{2}}}+\frac{a\sqrt{{a}^{2}+{b}^{2}}\mathrm{sin}x}{\sqrt{{a}^{2}+{b}^{2}}}$
$=b\mathrm{cos}x+a\mathrm{sin}x$
$\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B$
$\mathrm{cos}\left(\frac{\mathrm{arctan}a}{b}\right)=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$
$\mathrm{sin}\left(\frac{\mathrm{arctan}a}{b}\right)=\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}$

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