Question

if sin alpha = 15/17 ,cos bita = 12/13 , find the values of sin (alpha + bita) , cos (alpha - bita) and tan (alpha +bita) , where alpha , bita are +ve acute angles​

Answer 1

Answer:

sin(α+β) = 220 / 221

cos(α-β) = 21 /221

tan(α+β) = 220 / 21

Step-by-step explanation:

We are given that

sin(α) = 15/17

cos(β) = 12/13

we know that

          sin²(Ф) + cos²(Ф) = 1

When Ф = α

⇒        sin²(α) + cos²(α) = 1

Putting sin(α) = 15/17 we get

          (15/17)² + cos²(α) = 1

⇒                       cos²(α) = 1 -  (15/17)² = 1 - 225 / 289 = 64 /289

⇒                       cos²(α) = 64/289

Taking under roots on both sides we get

                           cos(α) = 8/17

Similarly when Ф = β then

            sin²(β) + cos²(β) = 1

By putting cos(β) = 12/13 we get

                           sin ²(β) = 1 -  (12/13)² = 1 - 144 / 169 = 25/169

Taking under root on both sides we get

                            sin (β) = 5/13

NOW

We know that

             sin(α+β) = sin(α)cos(β) + cos(α)sin(β)

Putting value we get

              sin(α+β) = (15/17)(12/13) + (8/17)(5/13) = (180 + 40) / 221 = 220 / 221

So          sin(α+β) = 220 / 221

NOW

We know that

           cos(α-β) = cos(α)cos(β) + sin(α)sin(β)

Putting values we get

           cos(α-β) = (8/17)(12/13) + (15/17)(5/13) = 171 / 221

So       cos(α-β) = 21 /221

NOW

we know that

tan(α+β) =  sin(α+β) /  cos(α+β)

And

cos(α+β) = cos(α)cos(β) - sin(α)sin(β)

Putting values we get

 cos(α+β) = (8/17)(12/13) - (15/17)(5/13) = 21 / 221

So

tan(α+β) =  sin(α+β) /  cos(α+β) = (220 / 221) / (21 / 221) = 220 / 21

Thus

    tan(α+β) = 220 / 21