cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

ncert 11th trigonometry, exercise 3.3

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

using trigonometry formula trigonometry identities

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

LHS = cos[(3pi/4)+x] - cos[(3pi/4)-x]

=2cos[2(3pi/4)/2] cos[2x/2]

=2cos[(3pi/4)] cosx

=2{-1 /(-sqrt(2)) } cosx   [cosx is negative in second quadrant]

= (-sqrt(2))sinx

=RHS

10. prove that

sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx

using trigonometry formula trigonometry identities
cosAcosB +sinAsinB = cos(A-B)

A =(n+2)x
B=(n+1)x


LHS =  sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x]

=cos[A-B] FORM

=cos[(n+2)x - (n+1)x ]

=cosx =RHS


6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)

using trigonometry formula trigonometry identities
cosAcosB - sinAsinB = cos(A+B)

A= [(pi/4)-x]
B= [(pi/4)-y]

LHS =  cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y]

= cos[A+B] form

 = cos[[(pi/4)-x]+[(pi/4)-y]]

=cos[ 2(pi/4)-(x+y)]

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

=sin[x+y] = RHS  using cos [ (pi/2)-A] =sinA

3.3

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

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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