exercise 3.3 ncert trigonometry

exercise 3.3 ncert trigonometry cbse 11th

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

using trigonometry formula trigonometry identities

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

(cosx)^2 - (sinx)^2 = cos2x in denominator


LHS = [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] introducing (-1) to change the
order in both numerator and denominator

=[sin3x - sinx] / [ (cosx)^2 - (sinx)^2 ]

= [2cos(4x/2) sin(2x/2)]  / [cos2x]

= [2cos(2x) sin(x)]  / [cos2x]

= 2 sinx = RHS





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

using trigonometry formula trigonometry identities

tan2x = { 2 tanx }  / { 1 - [(tanx)^2]}
Replace x with 2x

LHS = tan4x =  tan{2(2x)}

= { 2 tan2x }  / { 1 - [(tan2x)^2]} again using the same formula

={ 2[{ 2 tanx }  / { 1 - [(tanx)^2]}] }/ { 1 -[{ 2 tanx }  / { 1 - [(tanx)^2]}]^2 }

expand using identity for (a-b)^2 in denominator . . .

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




3.3

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


miscellaneous

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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