if cos A= (3/5) and tan B = ((12)/5) where A and B are reflex angles find without using tables,the value of a) sin (A − B) b) tan(A−B) c) cos (A + B).


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 39588 by Rio Mike last updated on 08/Jul/18

$i f c o s A = \frac{3}{5} a n d t a n B = \frac{12}{5}$ $w h e r e A a n d B a r e r e f l e x a n g l e s$ $f i n d w i t h o u t u s i n g t a b l e s, t h e$ $v a l u e o f$ $a) s i n (A - B) b) t a n (A - B)$ $c) c o s (A + B) .$
	Answered by MJS last updated on 08/Jul/18

	$I see two Pythagorean triples$ $3^{2} + 4^{2} = 5^{2} and 5^{2} + 12^{2} = 13^{2}$ $\cos α = \frac{3}{5} \Rightarrow \sin α = \frac{4}{5} \Rightarrow \tan α = \frac{4}{3}$ $\tan β = \frac{12}{5} \Rightarrow \cos β = \frac{5}{13} \Rightarrow \sin β = \frac{12}{13}$ $\sin (α - β) = \sin α \cos β - \cos α \sin β =$ $= \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} = - \frac{16}{65}$ $\tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β} =$ $= \frac{\frac{4}{3} - \frac{12}{5}}{1 + \frac{4}{3} \times \frac{12}{5}} = - \frac{16}{63}$ $\cos (α + β) = \cos α \cos β - \sin α \sin β =$ $= \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} = - \frac{33}{65}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum