cbse exercise 3.3 trigonometry problems
22. Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
using trigonometry formula trigonometry identities
tan(A+B) = [ tanA + tanB ] / [1-tanAtanB]
and then change [1 / tanx] = cotx
the angles used are x , 2x ,3x
So we connect them using the relation
3x = x + 2x
take tan on both sides
tan3x = tan( x + 2x )
tan3x = [ tanx + tan2x ] / [ 1 - tanx tan2x ]
rearranging
[ 1 - tanx tan2x ] (tan3x) = [ tanx + tan2x ]
simplifying
tan3x - tanx tan2x tan3x =[ tanx + tan2x ]
tan3x - tan2x - tanx = tanx tan2x tan3x
divide each term by tanx tan2x tan3x
{ 1/[tan2x tanx] } - { 1 / [tan2x tan3x] } -{ 1/ [tan3x tanx]} =1
using [1 / tanx] = cot x
we get,
cotx cot2x -cot2xcot3x-cot3xcotx = 1
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
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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