Ashleigh Mitchell

2022-04-08

Trigonometry problem $\frac{{\mathrm{sin}100}^{\circ }+{\mathrm{cos}70}^{\circ }}{{\mathrm{cos}80}^{\circ }-{\mathrm{cos}20}^{\circ }}$
I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:
$\frac{{\mathrm{sin}100}^{\circ }+{\mathrm{cos}70}^{\circ }}{{\mathrm{cos}80}^{\circ }-{\mathrm{cos}20}^{\circ }}=$
$\frac{\mathrm{sin}\left({90}^{\circ }+{10}^{\circ }\right)+\mathrm{cos}\left({60}^{\circ }+{10}^{\circ }\right)}{\mathrm{cos}\left({90}^{\circ }-{10}^{\circ }\right)-\mathrm{cos}\left({30}^{\circ }+{10}^{\circ }\right)}=$
$\frac{{\mathrm{sin}90}^{\circ }{\mathrm{cos}10}^{\circ }+{\mathrm{cos}90}^{\circ }{\mathrm{sin}10}^{\circ }+{\mathrm{cos}60}^{\circ }{\mathrm{cos}10}^{\circ }-{\mathrm{sin}60}^{\circ }{\mathrm{sin}10}^{\circ }}{{\mathrm{cos}90}^{\circ }{\mathrm{cos}10}^{\circ }+{\mathrm{sin}90}^{\circ }{\mathrm{sin}10}^{\circ }-{\mathrm{cos}30}^{\circ }{\mathrm{cos}10}^{\circ }+{\mathrm{sin}30}^{\circ }{\mathrm{sin}10}^{\circ }}=$
$\frac{{\mathrm{cos}10}^{\circ }+\frac{1}{2}{\mathrm{cos}10}^{\circ }-\frac{\sqrt{3}}{2}{\mathrm{sin}10}^{\circ }}{{\mathrm{sin}10}^{\circ }-\frac{\sqrt{3}}{2}{\mathrm{cos}10}^{\circ }+\frac{1}{2}{\mathrm{sin}10}^{\circ }}=$
$\frac{\frac{3}{2}{\mathrm{cos}10}^{\circ }-\frac{\sqrt{3}}{2}{\mathrm{sin}10}^{\circ }}{\frac{3}{2}{\mathrm{sin}10}^{\circ }-\frac{\sqrt{3}}{2}{\mathrm{cos}10}^{\circ }}$

analiticozuod

It is:
$\frac{{\mathrm{sin}100}^{\circ }+{\mathrm{cos}70}^{\circ }}{{\mathrm{cos}80}^{\circ }-{\mathrm{cos}20}^{\circ }}=\frac{{\mathrm{sin}80}^{\circ }+{\mathrm{sin}20}^{\circ }}{{\mathrm{cos}80}^{\circ }-{\mathrm{cos}20}^{\circ }}=\frac{2{\mathrm{sin}50}^{\circ }{\mathrm{cos}30}^{\circ }}{-2{\mathrm{sin}50}^{\circ }{\mathrm{sin}30}^{\circ }}=-\sqrt{3}$

seskew192atp

Given
$\frac{\mathrm{sin}100+\mathrm{cos}100}{\mathrm{cos}80-\mathrm{cos}20}$
Now to solve the denominator use the formula $\mathrm{cos}A-\mathrm{cos}B=-2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{sin}\left(\frac{A-B}{2}\right)$
$\mathrm{cos}80-\mathrm{cos}20=-2\mathrm{sin}\left(\frac{80+20}{20}\right)\mathrm{sin}\left(\frac{80-20}{2}\right)$
$=\frac{\mathrm{sin}100+\mathrm{cos}70}{-2\mathrm{sin}\left(\frac{80+20}{20}\right)\mathrm{sin}\left(\frac{80-20}{2}\right)}$
$=-\frac{\mathrm{sin}100+\mathrm{cos}70}{\mathrm{sin}50}$
Now to solve the numerator use the identity $\mathrm{sin}A+\mathrm{sin}B=2\mathrm{cos}\left(\frac{A-B}{2}\right)\mathrm{sin}\left(\frac{A+B}{2}\right)$
$\mathrm{sin}100+\mathrm{sin}20=2\mathrm{cos}\left(\frac{100-20}{2}\right)\mathrm{sin}\left(\frac{100+20}{2}\right)$
$=-\frac{2\mathrm{cos}\left(\frac{100-20}{2}\right)\mathrm{sin}\left(\frac{100+20}{2}\right)}{\mathrm{sin}50}$
$=-\frac{2\mathrm{cos}40\mathrm{sin}60}{\mathrm{sin}50}$
$=-\frac{2\mathrm{cos}40\cdot \frac{\sqrt{3}}{2}}{\mathrm{sin}50}$
$-\frac{\sqrt{3}\mathrm{sin}\left(90-40\right)}{\mathrm{sin}50}=-\sqrt{3}$
Therefore,
$\frac{\mathrm{sin}100+\mathrm{cos}100}{\mathrm{cos}80-\mathrm{cos}20}=-\sqrt{3}$

Do you have a similar question?