If a,b,c are in an Arithmetic Progression (AP), prove that
sinc−sinacosa−cosc=cotb.
I tried setting a,b,c as (a),(a+d),(a+2d) respectively as they are in an AP.It does not work at all. Is there any other method???
Answer
The trick is let a=A−d b=A and c=A+d.
sinc−sinacosa−cosc=sin(A+d)−sin(A−d)cos(A−d)−cos(A+d)
sinAcosd−cosAsind−sinAcosd−cosAsindcosAcosd−sinAsind−cosAcosd−sinAsind
Which simplifies to cosAsinA
cosAsinA=cotA
cotA=cotb
Quod erat demonstrandum.
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