Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

sinx + sin^2x +sin^3x =1


cos^6x-4cos^4x + 8 cos^2x = ?

Grade:12

3 Answers

vikas askiitian expert
509 Points
10 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 --- 1
2/3 sqrt(3)/2 -1/2 -sqrt(3) -sqrt(3)/3 -2 2 sqrt(3)/3
3/4 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                  /5
sin(x) (-1+sqrt[5])/4 sqrt(10-2 sqrt[5])/4
cos(x) sqrt(10+2 sqrt[5])/4 (1+sqrt[5])/4
tan(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.


[Back to Contents]

Bounds

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


[Back to Contents]

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
10 years ago

dear friend,

 

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

Soumyadeep
11 Points
4 years ago
Bro its simple Sin+sin3 = 1-sin2 Sin (1+sin2)=cos2Sin(2-cos2)=cos2Sq bth sides Sin2(4-4cos2+cos4)=cos41-cos2("""""""""""""")=cos44-4cos2+cos4-cos4-4cos2+4cos4-cos6=0Multiply -1cos6-4cos4+8cos2-4=0 :) These sums are very confusing try removing unwanted things instead of adding things.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free