show that ((sinα+sin3α+sin5α)/(cosα+cos3α+cos5α))= tan3α


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 52629 by scientist last updated on 10/Jan/19

$s h o w t h a t \frac{s i n α + s i n 3 α + s i n 5 α}{c o s α + c o s 3 α + c o s 5 α} = t a n 3 α$
	Answered by math1967 last updated on 10/Jan/19

	$\frac{s i n 3 α + 2 \sin 3 α . \cos 2 α}{\cos 3 α + 2 \cos 3 α \cos 2 α}$ $= \frac{\sin 3 α (1 + \cos 2 α)}{\cos 3 α (1 + \cos 2 α)} = t a n 3 α$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

	$\frac{s i n 5 α + s i n α + s i n 3 α}{c o s 5 α + c o s α + c o s 3 α}$ $= \frac{2 s i n 3 α . c o s 2 α + s i n 3 α}{2 c o s 3 α c o s 2 α + c o s 3 α}$ $= \frac{s i n 3 α (2 c o s 2 α + 1)}{c o s 3 α (2 c o s 2 α + 1)}$ $= t a n 3 α$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum