Question

If cos a =12/13
and sin b = 4/5, then find sin(a+b).

Answer 1

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Hey mate,

Seems like you're going through trouble solving Trignometric Identities

For thy convenience, here's a few rules one would need to know while solving 'em : 

→ sin ( A + B ) = ( sin A cos B + sin B cos A )
→ sin ( - A ) = - sin A
→ cos ( - A ) = - cos A

→ sin² A + cos² A = 1 

• Just knowing these would get you off with any trouble ! However, just in case, you'd like to read this answer :https://brainly.in/question/3221050
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Given : 
 → cos A = ( 12 / 13 )
 → sin B = ( 4 / 5 )

◘ Utilizing sin² X + cos² X = 1 , we get :
  → sin A = √( 1 - cos² A ) = ( 5 / 13 )
  → cos B = √( 1 - sin² B ) = ( 3 / 5 )

◘ Again from : sin ( A + B ) = ( sin A cos B + sin B cos A ) ;

We get : sin ( A + B ) = ( 5 / 13 )( 3 / 5 ) + ( 4 / 5 )( 12 / 13 ) 
                                  = ( 15 / 65 ) + ( 48 / 65 )
                                  = ( 63 / 65 )

► Result : sin ( A + B ) = ( 63 / 65 )
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Note : Quadratics say : √( 144 / 169 ) = ± ( 12 / 13 )
 
→ However, while solving for simple trigonometry, we avoid such possibilities 
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^_^ Hope it helps 

Answer 2

Given Equation is cosa = 12/13.

We know that sin^2a + cos^2a = 1

                         sin^2a + (12/13)^2 = 1

                         sin^2a + 144/169 = 1

                        sin^2a = 1 - 144/169

                        sin^2a = 25/169

                       sin a = 5/13.



Now,

cos^2b + sin^2b = 1

cos^2b + (4/5)^2 = 1

cos^2b + 16/25 = 1

cos^2b = 1 - 16/25

cos^2b = 9/25

cosb = 3/5.



We know that sin(a + b) = sin a cosb + sin b cosa

                                        = (5/13) * (3/5) + (4/5) * (12/13)

                                        $= \frac{3}{13} + \frac{48}{65}$

                                        $= \frac{3 * 5 + 48}{65}$

                                        $= \frac{63}{65}$



Hope this helps!