Question

sin6theta + cos6theta + 3sin2thetacos2theta=1 prove this

Answer 1

Hey mate!!!
Here is the answer to your query...

sin^6A + cos^6A + 3sin^2Acos^2A
= (sin^2)^3 + (cos^2)^3 + 3sin^2Acos^2A

[a^3 + b^3 = (a+b)(a^2 + b^2 - ab)]

=(sin^2 + cos^2)(sin^4 + cos^4 - sin^2 cos^2) + 3sin^2cos^2

(sin^2 + cos^2 = 1)

=(1)(sin^4 + cos^4 - sin^2cos^2) + 3sin^2 cos^2
= sin^4 + cos^4 - sin^2cos^2 + 3sin^2 cos^2
= sin^4 + cos^4 + 2sin^2cos^2

[a^2 + b^2 + 2ab = (a+b)^2]

= (sin^2 + cos^2)^2

=(1)^2

= 1 RHS

Hence proved...


Hope it helps!!!
If so, then let me know by marking this as the brainliest one...

Regards
Rhea

Answer 2

Answer:

sin^6A + cos^6A + 3sin^2Acos^2A

= (sin^2)^3 + (cos^2)^3 + 3sin^2Acos^2A

[a^3 + b^3 = (a+b)(a^2 + b^2 - ab)]

=(sin^2 + cos^2)(sin^4 + cos^4 - sin^2 cos^2) + 3sin^2cos^2

(sin^2 + cos^2 = 1)

=(1)(sin^4 + cos^4 - sin^2cos^2) + 3sin^2 cos^2

= sin^4 + cos^4 - sin^2cos^2 + 3sin^2 cos^2

= sin^4 + cos^4 + 2sin^2cos^2

[a^2 + b^2 + 2ab = (a+b)^2]

= (sin^2 + cos^2)^2

=(1)^2

= 1 RHS