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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
We are all IITians and here to help you in your IIT JEE preparation. All the best. If you like this answer please approve it.... win exciting gifts by answering the questions on Discussion Forum Sagar Singh B.Tech IIT Delhi
We are all IITians and here to help you in your IIT JEE preparation. All the best.
If you like this answer please approve it....
win exciting gifts by answering the questions on Discussion Forum
Sagar Singh
B.Tech IIT Delhi
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