miscellaneous trigonometry question 6 cbse ncert 11th mathematics

miscellaneous trigonometry question 6 cbse ncert 11th mathematics

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

using trigonometry formula trigonometry identities

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

LHS =

[(sin7x+sin5x )+(sin9x+sin3x)] / [(cos7x+cos5x)+(cos9x+cos3x)]

= [2sin(12x/2)cos(2x/2)+2sin(12x/2)cos(6x/2)] / [2cos(12x/2)cos(2x/2)+2cos(12x/2)cos(6x/2)]

= [2sin6xcosx +2sin6xcos3x] / [2cos6xcosx +2cos6xcos3x]

={2sin6x[cosx +cos3x]}/{2cos6x[cosx+cos3x]}

=sin6x/cos6x ,cancelling off the common factors

=tan6x











1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x


solution


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