trigonometry miscellaneous question 2 ncert cbse 11th

trigonometry miscellaneous question 2 ncert cbse 11th

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

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

LHS = {2sin[(3x+x)/2] cos[(3x-x)/2]}sinx +{-2sin[(3x+x)/2] sin[(3x-x)/2]} cosx

=2sin2xcosxsinx - 2sin2xsinxcosx = 0 = RHS


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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trigonometry problem on ncert cbse 11th miscellaneous

trigonometry problem on ncert cbse 11th miscellaneous exercise

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

using the formula 2cosxcosy = cos(x+y)+cos(x-y)
trigonometry identities

LHS= cos[(pi/13) +(9pi/13)]+cos[(pi/13) -(9pi/13)]+cos (3pi/13)+cos(5pi/13)

= cos (10pi/13) + cos(-8pi/13) +cos (3pi/13)+cos(5pi/13)

= cos (10pi/13) + cos(8pi/13) +cos (3pi/13)+cos(5pi/13) because cos(-x)=cosx

= cos[pi - (3pi/13)] +cos[pi - (5pi/13)]+cos (3pi/13)+cos(5pi/13)

= -cos (3pi/13)- cos(5pi/13)+cos (3pi/13)+cos(5pi/13) because cos(pi-x)= -cosx

= 0


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