ncert 11th cbse trigonometry exercise 3.3 question16

ncert 11th cbse trigonometry exercise 3.3 question16


16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x

using trigonometry formula trigonometry identities

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

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

LHS = [cos9x -cos5x] / [sin17x - sin3x ]

={ -2sin(14x/2) sin(4x/2) } / { 2cos(20x/2)sin(14x/2) }

= {-2sin7xsin2x} / {2cos10x sin7x}

=  (-sin2x) /cos10x

=RHS

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]

using trigonometry formula trigonometry identities

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

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

LHS = [sinx -siny] / [cosx +cosy]

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

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

=tan[(x-y)/2]

=RHS




3.3

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution


19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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exercise 3.3 ncert trigonometry question 19

exercise 3.3 ncert trigonometry question 17 and 19

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x

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 =[sin5x + sin3x] / [cos5x+cos3x]

=[2sin(8x/2)cos(2x/2)] / [2cos(8x/2)cos(2x/2)]

= [2sin4xcosx]/[2cos4xcosx]

=sin4x/cos4x

=tan4x


19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x

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]

after rearrangement, after interchanging the terms to make the bigger angle first

LHS = [sinx + sin3x] / [cosx+cos3x]

=[sin3x + sinx] / [cos3x+cosx]

=[2sin(4x/2)cos(2x/2)] / [2cos(4x/2)cos(2x/2)]

=[2sin2xcosx] / [2cos2xcosx]

=sin2x/cos2x

=tan2x

=RHS

3.3

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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miscellaneous trigonometry question 7 cbse ncert 11th mathematics

miscellaneous trigonometry question 7 cbse ncert 11th mathematics

7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)


using trigonometry formula trigonometry identities

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

sinx = 2sin(x/2)cos(x/2)

and then

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]
after taking out factor

LHS=  (sin3x+sin2x) - sinx

=2sin(5x/2)cos(x/2) - 2sin(x/2)cos(x/2)

=2cos(x/2) [sin(5x/2) - sin(x/2)]

= 2cos(x/2) [ 2cos((6x/2)/2)sin((4x/2)/2) ]

=2cos(x/2) [ 2cos(3x/2)sin(x) ]


=4sin(x)cos(x/2)cos(3x/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

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

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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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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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



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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

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