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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11th cbse trigonometry miscellaneous exercise question 3

11th cbse trigonometry miscellaneous exercise question 3

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

using the trigonometry formulae
cosx + cosy = 2cos[(x+y)/2] cos[(x-y)/2]
sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS =  { 2cos[(x+y)/2] cos[(x-y)/2] }^2 + {2cos[(x+y)/2] sin [(x-y)/2]}^2

taking common factor out

= 4{ cos[(x+y)/2] }^2 [{ cos[(x-y)/2] }^2 + { sin[(x-y)/2] }^2 ]

= 4{ cos[(x+y)/2] }^2 [1]

=4{ cos[(x+y)/2] }^2 = RHS.


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

using trigonometry identity

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


LHS =  {- 2sin[(x+y)/2] sin[(x-y)/2] }^2 + {2cos[(x+y)/2] sin [(x-y)/2]}^2

= 4 { sin[(x-y)/2] }^2 [{ sin[(x+y)/2] }^2 + { cos[(x-y)/2] }^2 ]

= 4 { sin[(x-y)/2] }^2 [ 1 ]

= 4 { sin[(x-y)/2] }^2



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



disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work