Question

If Sin a= 12/13 in quad. 2 , cos b=-4/5 quad 3, how do I find sin (a+b) ?

Answer 1

Answer:

Therefore

$\sin (a+b) =-\dfrac{23}{65}$

Step-by-step explanation:

Given:

$\sin a =\dfrac{12}{13}\\\\\cos b=-\dfrac{4}{5}$

To Find:

$\sin (a+b)= ?$

Solution:

Using identity,

$\sin^{2}a+\cos^{2}a=1$

Now Substitute Sin a we get

As 'a' is in Second Quadrant Cos a is Negative

Similarly,

$\sin^{2}b+\cos^{2}b=1$

Now Substitute Cos b we get

$(\dfrac{-4}{5})^{2}+\sin^{2}b=1\\\\\sin^{2}b=1-\dfrac{16}{25}=\dfrac{9}{25}\\\\\sin b=\pm\sqrt{\dfrac{9}{25}}\\\\\sin b=-\dfrac{3}{5}$

As 'b' is in Third Quadrant Sin b is Negative

Now Using Identity

$\sin (a+b) = \sin a.\cos b + \cos a.\sin b$

Substituting the values we get

$\sin (a+b) =\dfrac{12}{13}\times -\dfrac{4}{5}+\dfrac{-5}{13}\times -\dfrac{3}{5}$

$\sin (a+b) =\dfrac{-48+15}{13\times 5}$

$\sin (a+b) =-\dfrac{23}{65}$

Therefore

$\sin (a+b) =-\dfrac{23}{65}$