Trigonometry-

Question Number 120614 by TANMAY PANACEA last updated on 01/Nov/20

T r i g o n o m e t r y

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

t a n \frac{π}{9} + 4 s i n \frac{π}{9}

= \frac{1}{c o s 20 °} (s i n 20 ° + 4 c o s 20 s i n 20 °)

= \frac{1}{c o s 20 °} (s i n 20 ° + 2 s i n 40 °)

= \frac{1}{c o s 20 °} (2 s i n 30 ° c o s 10 ° + s i n 40 °)

= \frac{1}{c o s 20 °} (s i n 80 ° + s i n 40 °)

= \frac{1}{c o s 20 °} (2 s i n 60 ° c o s 20 °)

= \sqrt{3}

Commented by Dwaipayan Shikari last updated on 01/Nov/20

t a n (\frac{2 π}{9}) - 4 s i n (\frac{2 π}{9})

= \frac{1}{c o s 40 °} (s i n 40 ° - 4 s i n 40 ° c o s 40 °

= \frac{1}{c o s 40 °} (s i n 40 ° - 2 s i n 80 °)

= - \frac{1}{c o s 40 °} (2 c o s 60 ° s i n 20 ° + s i n 80 °)

= - \frac{1}{c o s 40 °} (2 s i n 50 c o s 30)

= - \sqrt{3}

Commented by TANMAY PANACEA last updated on 01/Nov/20

t h a n k y o u

Commented by TANMAY PANACEA last updated on 01/Nov/20

t h a n k y o u