# sinx + sin^2x +sin^3x =1cos^6x-4cos^4x + 8 cos^2x = ?

509 Points
13 years ago

### Definitions of Trigonometric Functions

Draw a unit circle with center O. Let a central angle with initial side OP and terminal side OQ contain x radians (that is, the arc PQ has length x). Drop a perpendicular from Q to OP meeting it at R. Then OR = cos(x) and RQ = sin(x). If those directed line segments are up or to the right, the lengths are positive. If they are down or to the left, the lengths are negative.

Values at special angles:

`  x      sin(x)     cos(x)      tan(x)      cot(x)       sec(x)       csc(x)     0        0          1           0          ---           1           --- /6      1/2      sqrt(3)/2   sqrt(3)/3   sqrt(3)     2 sqrt(3)/3      2 /4   sqrt(2)/2   sqrt(2)/2      1           1         sqrt(2)      sqrt(2) /3   sqrt(3)/2     1/2       sqrt(3)     sqrt(3)/3       2       2 sqrt(3)/3 /2       1          0          ---          0           ---           12/3   sqrt(3)/2    -1/2      -sqrt(3)    -sqrt(3)/3      -2       2 sqrt(3)/33/4   sqrt(2)/2  -sqrt(2)/2     -1          -1        -sqrt(2)      sqrt(2)5/6      1/2     -sqrt(3)/2  -sqrt(3)/3  -sqrt(3)    -2 sqrt(3)/3      2          0         -1           0          ---          -1           ---`

More values at special angles:

`  x              /10                  /5sin(x)      (-1+sqrt[5])/4      sqrt(10-2 sqrt[5])/4cos(x)   sqrt(10+2 sqrt[5])/4      (1+sqrt[5])/4tan(x)    sqrt(1-2/sqrt[5])      sqrt(5-2 sqrt[5])cot(x)    sqrt(5+2 sqrt[5])      sqrt(1+2/sqrt[5])sec(x)    sqrt(2-2/sqrt[5])         -1+sqrt[5]csc(x)        1+sqrt[5]          sqrt(2+2/sqrt[5])`

Use the above values and the identities below to obtain values of trigonometric functions of the following multiples of /10:

`3/10 = /2 - /5,2/5  = /2 - /10,3/5  = /2 + /10,7/10 = /2 + /5,4/5  =    - /5,9/10 =    - /10.`

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### Bounds

`   |sin(x)| <= 1,   |cos(x)| <= 1,   |sec(x)| >= 1,   |csc(x)| >= 1.`

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### Identities

`   sec(x) = 1/cos(x),   csc(x) = 1/sin(x),   cot(x) = 1/tan(x),   tan(x) = sin(x)/cos(x),   cot(x) = cos(x)/sin(x).   sin(-x) = -sin(x),   cos(-x) = cos(x),   tan(-x) = -tan(x),   cot(-x) = -cot(x),   sec(-x) = sec(x),   csc(-x) = -csc(x).   sin(/2-x) = cos(x),   cos(/2-x) = sin(x),   tan(/2-x) = cot(x),   cot(/2-x) = tan(x),   sec(/2-x) = csc(x),   csc(/2-x) = sec(x).   sin(/2+x) = cos(x),   cos(/2+x) = -sin(x),   tan(/2+x) = -cot(x),   cot(/2+x) = -tan(x),   sec(/2+x) = -csc(x),   csc(/2+x) = sec(x).   sin(-x) = sin(x),   cos(-x) = -cos(x),   tan(-x) = -tan(x),   cot(-x) = -cot(x),   sec(-x) = -sec(x),   csc(-x) = csc(x).   sin(+x) = -sin(x),   cos(+x) = -cos(x),   tan(+x) = tan(x),   cot(+x) = cot(x),   sec(+x) = -sec(x),   csc(+x) = -csc(x).   sin(2+x) = sin(x),   cos(2+x) = cos(x),   tan(2+x) = tan(x),   cot(2+x) = cot(x),   sec(2+x) = sec(x),   csc(2+x) = csc(x).   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,`
aku -- kumar
38 Points
13 years ago

dear friend,

kindly send your address, so that i'll send the solutions by post