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Question

Determine the smallest positive value of x (in degrees) for which tan (x+100)=tan(x+50)tan x tan (x50).

Prove that cos 7xcos 8x1+2 cos 5x=cos 2xcos 3x


Solution

We have,

tan (x+100)=tan(x+50)tan x tan (x50)

   tan (x+100)tan (x50)=tan(x+50)tan x

  sin (x+100)cos(x50)cos(x+100)sin(x50)=sin(x+50)sin xcos(x+50)cos x

Applying componendo and dividendo, we get

sin(x+100)cos(x50)+cos(x+100)sin(x50)sin(x+100)cos(x50)cos(x+100)sin(x50)

=sin(x+50)sin x+cos(x+50)cos xsin(x+50)sin xcos(x+50)cos x

   sin(x+100+x50)sin(x+100x+50)=cos(x+50x)cos(x+50+x)

[  sin(A±B)=sin A cos B±cos A sin Band cos(A±B)=cos A cos Bsin A sin B]

   ssin (2x+50)sin 150=cos 50cos (2x+50)

   sin(2x+50) cos (2x+50)=sin 150 cos 50

  2 sin (2x+50)cos (2x+50)=2×12 cos 50

[    sin 150=12]

    sin (4x+100)=cos 50

[ 2 sin A cos A=sin 2A]

     sin(4x+100)=sin(27050)

     4x+100=27050

     4x=27050100, 4x=270150

     4x=120

     x=30

We have, LHS=cos 7xcos 8x1+2 cos 5x

Multiply numerator and denominator by 2 sin5x2, we get

LHS=2 sin5x2 cos 7x2 sin 5x2 cos 8x2 sin 5x2+4 sin 5x2 cos 5x

=(sin19x2sin9x2)(sin 21x2sin 11x2)2 sin 5x2+2 sin15x22 sin 5x2

[ 2 sin A ccos B  = sin (A + B) + sin (A - B)]

=(sin19x2+sin11x2)(sin9x2+sin21x2)2 ssin 15x2

=2 sin15x2 cos 2x2 sin 15x2 cos 3x2 sin 15x2

[   sin C+sin D=2 sinC+D2 cos CD2]

= cos 2x - cos 3x = RHS   Hence proved.

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