Inyalan0

2021-12-26

What is the value of $\frac{1}{{\mathrm{cos}18}^{\circ }{\mathrm{sin}9}^{\circ }}+\frac{1}{{\mathrm{cos}18}^{\circ }{\mathrm{cos}9}^{\circ }}$?

usaho4w

Hint:
Multiply the num/denom of the first fraction by $2\mathrm{cos}\left({9}^{\circ }\right)$ and the num/denom of the second fraction by $2\mathrm{sin}\left({9}^{\circ }\right)$ and use $2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$
you will get
$\frac{4\left({\mathrm{cos}9}^{\circ }+{\mathrm{sin}9}^{\circ }\right)}{{\mathrm{sin}36}^{\circ }}$
Finish by using the sum formulas for
${\mathrm{cos}9}^{\circ }+{\mathrm{sin}9}^{\circ }={\mathrm{sin}81}^{\circ }+{\mathrm{sin}9}^{\circ }=2{\mathrm{sin}45}^{\circ }{\mathrm{cos}36}^{\circ }$

Mary Herrera

hint
$a\mathrm{cos}x+b\mathrm{sin}x=$
$\sqrt{{a}^{2}+{b}^{2}}\mathrm{cos}\left(\theta -x\right)$
if a=b=1 then
$\theta ={45}^{\circ }$
and
${\mathrm{cos}9}^{\circ }+{\mathrm{sin}9}^{\circ }=\sqrt{2}\mathrm{cos}\left(45-9\right)$

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