Question

if sinx ,cosx are the roots of t^2 -pt + q =0 then show that sin^3+cos ^3=p ^3_3pq

Answer 1

Given: sinx ,cosx are the roots of t^2 -pt + q =0

To find: Prove that sin^3x + cos^3x =  (p/q)^3 - 3p

Solution:

• Now we have given the equation t^2 -pt + q =0

• The roots are: sinx ,cosx

• So sum of roots is:

                 sin x + cos x = - (-p)/q .............(i)

• Product of roots is:

                 sin x cos x = q .....................(ii)

• Now we have given sin^3x + cos^3x =  (p/q)^3 - 3p

• Consider LHS, we have:

                 sin^3x + cos^3x = (sin x + cos x) (sin^2 x + cos^2 x - sin x cos x)

• Now adding and subtracting 2 sin x cos x in second bracket, we get:

                 sin^3x + cos^3x = (sin x + cos x) (sin^2 x + cos^2 x - sin x cos x + 2sin x cos x - 2sin x cos x )

• Rearranging the terms, we get:

                 sin^3x + cos^3x = (sin x+cos x) (sin^2 x+cos^2 x+2sin x cosx-3sinxcosx )

                 sin^3x + cos^3x = (sin x+cos x) ( (sin x + cos x)^2  - 3sin x cos x )

• Now putting (i) and (ii) in above equation, we get:

                 sin^3x + cos^3x = (p/q) ((p/q)^2 - 3(q))

                 sin^3x + cos^3x = (p/q)^3 - 3q(p/q)

                sin^3x + cos^3x = (p/q)^3 - 3p

Answer:

        So we proved that sin^3x + cos^3x =  (p/q)^3 - 3p

Answer 2

Step-by-step explanation:

use (a+b)^3 formula then u can get the correct proof of it. hope u got the answer