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

Beginner2021-12-27Added 39 answers

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$\mathrm{sin}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\mathrm{cos}$

$\mathrm{cos}9}^{\circ}+{\mathrm{sin}9}^{\circ}={\mathrm{sin}81}^{\circ}+{\mathrm{sin}9}^{\circ}=2{\mathrm{sin}45}^{\circ}{\mathrm{cos}36}^{\circ$

Multiply the num/denom of the first fraction by

you will get

Finish by using the sum formulas for

Mary Herrera

Beginner2021-12-28Added 37 answers

hint

$a\mathrm{cos}x+b\mathrm{sin}x=$

$\sqrt{{a}^{2}+{b}^{2}}\mathrm{cos}(\theta -x)$

if a=b=1 then

$\theta ={45}^{\circ}$

and

${\mathrm{cos}9}^{\circ}+{\mathrm{sin}9}^{\circ}=\sqrt{2}\mathrm{cos}(45-9)$

if a=b=1 then

and

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