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



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