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kiran venkatesh Grade: 11
`        prove that:sinAcosB+cosAsinB=sin(A+B)sin(A-B)`
7 years ago

## Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
```										Dear kiran,
Given the functions (sinα, cosα, sinβ and cos β), we seek a formula that expresses sin(α+β).  ABC which has an angle α ACD which  "   "    "  β The long side ("hypotenuse') of ACD is AD=R. Therefore   DC = R sin β AC = R cos β Similarly  BC = AC sin α = R cos β sin α AB = AC cos α = R cos β cos α  The triangle ADF is right-angled and has the angle (α+β). Therefore  R sin (α+β) = DF R cos (α+β) = AF   Start by deriving the sine: R sin (α+β) = DF  =  EF + DE  =  BC + DE   Note in the drawing the two head-to-head angles marked with double  lines: like all such angles, they must be equal. Each of them is one of  the two sharp ("acute") angles in its own right-angled triangle. Since  the sharp angles in such a triangle add up to 90 degrees, the other two  sharp angles must be equal. This justifies marking the angle near D as  α, as drawn in the figure. In the right-angled triangle CED   DE = DC cos α = R sin β cos α EC = DC sin α = R sin β sin α Earlier it was already shown that BC = R cos β sin α AB = R cos β cos α Therefore R sin (α+β)  =  BC+DE  =  R cos β sin α + R sin β cos α  Cancelling R we have    sin (α+β)  =  sin α cos β  + cos α sin β

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7 years ago
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