Question

Two answers are given , i am getting only one answer

Answer 1

\[\large Question\]If sin a = 3/5 , cos b = 5/13 , then find sin(a+b)

Answer 2

\[\huge My sollution\]

Answer 3

You are probably getting only one answer because you would not have considered the \(\pm\) signs:

\[\large{\cos a = \pm \cfrac{4}{5}}\]

\[\large{\sin b = \pm \cfrac{12}{13}}\]

Answer 4

\[\huge \sin(a+b)= sina cosb + cosasinb\]
I substituted the values in the identity , and got
63/65

Answer 5

Yeah you forgot the \(\pm\) signs

Answer 6

I am not getting where i missed out something

Answer 7

You have not considered the negative values of cos a and sin b

Answer 8

why will they be negative

Answer 9

Because:

the correct relationship is this:

\[\large{\sin x = \pm \sqrt{1-\cos^2 x}}\]

Answer 10

i can use that , but , i should get two values using my method also

Answer 11

no because you assumed that your triangle lies in a particular quadrant

Answer 12

oh, that slipped of my mind thanks!

Answer 13

yw :)

Answer 14

I am getting answers as 63/65 and
-38/65

in the book it is given 63/65 and -33/65

Answer 15

I think the reason is from sina = sin(90 +a) , so that you have to separate it into 2 cases: one is your solution, and other is sin (90+a)
and we have sin ((90+a)+ b) = sin(90+a) cos b + sinb cos (90+a)
but cos (90+a) = -cos a , you can get -33/65 from it.