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Trigonometry> Prove that [cosx – sinx][cos2x – sin2x] =...

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

Kawal , 11 Years ago

Grade 11

3 Answers

Jitender Singh

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

Approved

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