Question

if cos x=3/5 and cos y= -24/25 ,where 270 <x<360 and 180 < y< 270,find the value of (i) sin(x+y) (ii) cos (x-y) (iii).tan(x+y).


plz help me guys..

Answer 1

^_^ TRIGONOMETRY •.^

Steps:
1) When
$\cos(x) = \frac{3}{5}$
and.
$\frac{3\pi}{4} < x < 2\pi$
then.
$\sin(x) = \frac{ - 4}{5 } \\ \tan(x) = \frac{ - 4}{3}$

2) When
$\cos(y) = \frac{ - 24}{25}$
and
$\pi< y < \frac{3\pi}{4}$
then
$\sin(y) = \frac{ - 7}{25} \\ \tan(y) = \frac{7}{24}$

3) Now to the question,

$\sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y) \\ = > \frac{ - 4}{5} \times \frac{ - 24}{25} + \frac{3}{5} \times \frac{ - 7}{25} \\ = > \frac{75}{125} = \frac{3}{5}$

4)
$\cos(x - y) = \cos(x) \cos(y) + \sin(x) \sin(y) \\ = > \frac{ 3}{5} \times \frac{ - 24}{25} + (\frac{ - 4}{5}) \times ( \frac{ - 7}{25} ) \\ = > \frac{ - 44 }{125}$
5)
$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) } \\ = > \frac{ \frac{ - 4}{3} + \frac{7}{24} }{1 - ( \frac{ - 4}{3}) \times \frac{7}{24} } \\ = > \frac{ - 75}{100} = \frac{ - 3}{4}$
^_^ Hope, you Enjoyed reading ^.^
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