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Prove that [cosx – sinx][cos2x – sin2x] = cosx – sin3x

Prove that
[cosx – sinx][cos2x – sin2x] = cosx – sin3x

Grade:11

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

(cosx - sinx)(cos2x - sin2x) = (cosx - sin3x)
RHS = cosx - sin3x
RHS = cosx - sin(x+2x)
RHS = cosx - (sinx.cos2x + cosx.sin2x)
RHS = cosx - sinx.cos2x - cosx.sin2x
LHS = (cosx - sinx)(cos2x - sin2x)
LHS = cosx.cos2x-cosx.sin2x - sinx.cos2x+sinx.sin2x
LHS = cosx.cos2x+sinx.sin2x -cosx.sin2x - sinx.cos2x
LHS = cosx.(1-2sin^{2}x)+sinx.2sinx.cosx -cosx.sin2x - sinx.cos2x
LHS = cosx-2sin^{2}x.cosx+2sin^{2}x.cosx -cosx.sin2x - sinx.cos2x
LHS = cosx -cosx.sin2x - sinx.cos2x
LHS = RHS
Hence Proved
laxman
18 Points
6 years ago
wqwqwqwq
laxman
18 Points
6 years ago
the value of cos6 sin24 cos72
 

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