Question

plz answer this fast its hard

Answer 1

keep SinA that side in subtraction with cos2A. Sin3A =cos2A-SinA....now take 1 common from sin3A.....and as we know that (cos2A=1-SinA).....so keep it in the equation it will become SinA(Sin2A)=1-SinA-SinA.....now u can solve it further.....and take both side as Solve RHS along with LHS...u will get it bro....thanks

Answer 2

Hey there,

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 identity and you'll prove that it will equal 4.