\(\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}{\sum_{{{r}={0}}}^{{4}}}{\tan{{\left({\frac{{\pi}}{{{15}}}}+{\frac{{{r}\pi}}{{{5}}}}\right)}}}=?\) Then evaluation of k solution i try \(\displaystyle{\tan{{\left({12}^{\circ}\right)}}}+{\tan{{\left({48}^{\circ}\right)}}}+{\tan{{\left({84}^{\circ}\right)}}}+{\tan{{\left({120}^{\circ}\right)}}}+{\tan{{\left({156}^{\circ}\right)}}}\)

dotzis16xd

dotzis16xd

Answered question

2022-03-28

13r=04tan(π15+rπ5)=?
Then evaluation of k
solution i try
tan(12)+tan(48)+tan(84)+tan(120)+tan(156)
=tan(12)+tan(48)+tan(84)+tan(156)+3
=sin(60)cos(12)cos(48)+sin(240)cos(84)cos(156)3

Answer & Explanation

Alexzander Evans

Alexzander Evans

Beginner2022-03-29Added 9 answers

S=tan(12)+tan(48)+tan(84)+tan(120)+tan(156)
S=tan(12)+tan(48)+tan(84)+tan(156)+3
S=sin(60)cos(12)cos(48)+sin(240)cos(84)cos(156)3
S=3cos(60)+cos(36)3cos(240)+cos(72)3
use cos(36)=5+14 and cos(72)=sin(18)=514
=43[64]3=53

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