cbse 11th trigonometry exercise 3.3

cbse 11th trigonometry exercise 3.3

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

using trigonometry formula trigonometry identities


use the formula cos2x = 1-2(sinx)^2 with x replaced by (2x)
and then use
sin2x = 2 sinx cosx


LHS = cos4x = cos[2(2x)]

= 1-2[ sin(2x) ]^2

= 1 - 2[ 2 sinx cosx ]^2

= 1 -8[(sinx)^2][(cosx)^2] = RHS


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

using trigonometry formula trigonometry identities

cos3x = 4[ (cosx)^3] - 3[ cosx ] with x replaced by 2x

and then use  cos2x = 2 [(cosx)^2] - 1

cos6x = cos[3(2x)] =  4[ (cos2x)^3] - 3[ cos2x ]

=4{ [ 2 [(cosx)^2] - 1 ] ^3} - 3{ 2 [(cosx)^2] - 1 }

using the identity for (a-b)^3 = (a^3) -3(a^2)b +3 a(b^2) - (b^3)

=4{ 8  [(cosx)^6]  - 12[(cosx)^4] + 6[(cosx)^2] - 1 } -3{ 2 [(cosx)^2] - 1 }

= 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1  on simplification




3.3

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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