Question

prove that sin5x-2sin3x+sinx/cos5x-cosx​

Answer 1

Answer:





Join / Login

Question



Prove that =cos5x−cosxsin5x−2sin3x+sinx=tanx

Medium

Open in App

Solution



Verified by Toppr

Formula:

sinC−sinD=2cos(2C+D)sin(2C−D)

cosC−cosD=−2sin(2C+D)sin(2C−D)

LHS=cos5x−cosxsin5x−2sin3x+sinx

=cos5x−cosx(sin5x−sin3x)−(sin3x−sinx)

=−2sin3xsin2x2cos4xsinx−2cos2xsinx

=−2sin3xsin2x2sinx(cos4x−cos2x)

=−2sin3x(2sinxcosx)

Answer 2

ㅤ

Prove that -

$\frac{sin \: 5x - 2sin \: 3x \: + sin \: x}{cos \: 5x - cos \: x} = tan \: x$

Formula -

$sin( - sin \: d = 2 \: cos( \frac{c + d}{2} )sin( \frac{c - d}{2} )$

$cos \: c - cos \: d = - 2sin \frac{c + d}{2}sin \frac{c - d}{2}$

LHS -

$= \frac{sin \: 5x -2 sin \: 3x + sin \: x}{cos \: 5x - cos \: x}$

$= \frac{(sin \: \: 5x \: - sin \: 3x) - (sin \: 3x - sin \: x)}{cos \: 5x - cos \: x}$

$= \frac{2 \: cos \: 4 \:x \: sin \: x - 2 \: cos \: 2x \: sin \: x}{ - 2 \: sin \: 3x \: sin \: 2x}$

$= \frac{2 \: sin \: x \: ( \: - 2 \: sin \: 3x \: sin \: x)}{ - 2 \: sin \: 3x \: (2 \: sin \: x \: cos \: x)}$

$= \frac{sin \: x}{cos \: x}$

= tan x = RHS

LHS = RHS

━━━━━━━━━━━━━━━━━━━〢〢〢〢