miscellaneous trigonometry question 5 cbse ncert 11th mathematics

miscellaneous trigonometry question 5 cbse ncert 11th mathematics

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

using trigonometry formula trigonometry identities

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

after rearranging

LHS

= (sin 7x + sin x) + (sin5x + sin3x)

= 2 sin(8x/2)cos(6x/2) + 2 sin(8x/2)cos(2x/2)

=2 sin4x cos3x  + 2 sin4x cos x

= 2sin4x [cos 3x + cos x]  taking common factor

= 2sin4x [2 cos(4x/2) cos(2x/2)] using formula cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

=2sin4x [2 cos(2x) cos(x)]

= 4cosx cos2x sin 4x

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



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