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 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}

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