Does anyone know how to solve these trigonometric equations and identities??

Question

1. sin^4x + cos^4x=______________

*the symbol(^) means to the power of 4

2. sin(a+B) + cos(a-B)/cos(a+B) - sin(a-B)=__________

*a means alpha and B means beTa

...im really poor in solving trig...sO please help me!thankz!

Answer 1

This should help -

   sin2(x) + cos2(x) = 1,

   tan2(x) + 1 = sec2(x),

   1 + cot2(x) = csc2(x).

   sin(x+y) = sin(x)cos(y) + cos(x)sin(y),

   cos(x+y) = cos(x)cos(y) - sin(x)sin(y),

   tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)],

   cot(x+y) = [cot(x)cot(y)-1]/[cot(x)+cot(y)].

   sin(x-y) = sin(x)cos(y) - cos(x)sin(y),

   cos(x-y) = cos(x)cos(y) + sin(x)sin(y),

   tan(x-y) = [tan(x)-tan(y)]/[1+tan(x)tan(y)],

   cot(x-y) = [cot(x)cot(y)+1]/[cot(y)-cot(x)].

   sin(2x) = 2 sin(x)cos(x),

   cos(2x) = cos2(x) - sin2(x),

            = 2 cos2(x) - 1,

            = 1 - 2 sin2(x),

   tan(2x) = [2 tan(x)]/[1-tan2(x)],

   cot(2x) = [cot2(x)-1]/[2 cot(x)].

   |sin(x/2)| = sqrt([1-cos(x)]/2),

  

   |cos(x/2)| = sqrt([1+cos(x)]/2),

  

   |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]),

  

   tan(x/2) = [1-cos(x)]/sin(x),

            = sin(x)/[1+cos(x)].

            

   sin(3x) = 3 sin(x) - 4 sin3(x),

   cos(3x) = 4 cos3(x) - 3 cos(x),

   tan(3x) = [3 tan(x)-tan3(x)]/[1-3 tan2(x)].

   sin(4x) = 4 sin(x)cos(x)[2 cos2(x)-1],

   cos(4x) = 8 cos4(x) - 8 cos2(x) + 1.

   sin(5x) = 5 sin(x) - 20 sin3(x) + 16 sin5(x),

   cos(5x) = 16 cos5(x) - 20 cos3(x) + 5 cos(x).

   sin(6x) = 2 sin(x)cos(x)[16 cos4(x) - 16 cos2(x) + 3],

   cos(6x) = 32 cos6(x) - 48 cos4(x) + 18 cos2(x) - 1.

   sin(nx) = 2 sin([n-1]x)cos(x) - sin([n-2]x),

   cos(nx) = 2 cos([n-1]x)cos(x) - cos([n-2]x),

   tan(nx) = (tan[(n-1)x]+tan[x])/(1-tan[(n-1)x]tan[x...

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

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

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

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

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

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

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

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

   tan(x) + tan(y) = sin(x+y)/[cos(x)cos(y)],

   tan(x) - tan(y) = sin(x-y)/[cos(x)cos(y)],

   cot(x) + cot(y) = sin(x+y)/[sin(x)sin(y)],

   cot(x) - cot(y) = -sin(x-y)/[sin(x)sin(y)].

   [sin(x)+sin(y)]/[cos(x)+cos(y)] = tan[(x+y)/2],

   [sin(x)-sin(y)]/[cos(x)+cos(y)] = tan[(x-y)/2],

   [sin(x)+sin(y)]/[cos(x)-cos(y)] = -cot[(x-y)/2],

   [sin(x)-sin(y)]/[cos(x)-cos(y)] = -cot[(x+y)/2],

   [sin(x)+sin(y)]/[sin(x)-sin(y)] = tan[(x+y)/2]/tan[(x-y)/2].

   sin2(x) - sin2(y) = sin(x+y)sin(x-y),

   cos2(x) - cos2(y) = -sin(x+y)sin(x-y),

   cos2(x) - sin2(y) = cos(x+y)cos(x-y).

  

   sin2(x) = (1 - cos[2x])/2,

   cos2(x) = (1 + cos[2x])/2,

   tan2(x) = (1 - cos[2x])/(1 + cos[2x]),

   sin3(x) = (3 sin[x] - sin[3x])/4,

   cos3(x) = (3 cos[x] + cos[3x])/4,

   sin4(x) = (3 - 4 cos[2x] + cos[4x])/8,

   cos4(x) = (3 + 4 cos[2x] + cos[4x])/8,

   sin5(x) = (10 sin[x] - 5 sin[3x] + sin[5x])/16,

   cos5(x) = (10 cos[x] + 5 cos[3x] + cos[5x])/16,

   sin6(x) = (10 - 15 cos[2x] + 6 cos[4x] - cos[6x])/32,

   cos6(x) = (10 + 15 cos[2x] + 6 cos[4x] + cos[6x])/32,

Answer 2

Sin^4x= (Sin^2x)^2= (1-cos^2x)^2= 1-2Cos^2x+Cos^4x

Sin^4x+Cos^4x=  1-2 Cos^2x+2 Cos^4 x

= 1-2Cos^2x( 1-Cos^2x)

=1-2Cos^2x.Sin^2x

=1- (1/2)(2Sinx.Cosx)^2= 1- (1/2)Sin^2(2x)

(2)

Nr = (Sin A Cos B + Cos A Sin B)(Cos A Cos B +Sin A.Sin B)

=Sin A.Cos A.Cos^B+ Sin^2ASinB.CosB+Cos^2A SinB CosB+Sin A.Cos A.Sin^B

= Sin A Cos A( Sin^B+Cos^B) + SinB Cos B( Cos^2A+Sin^2A)= Sin A Cos A + Sin B CosB

In similar lines

Dr = Sin A Cos A- Sin B Cos B

Answer =

(Sin A Cos A + Sin B CosB)/(Sin A Cos A- Sin B Cos B)