Question

8. Prove that '3(sin theta-cos theta)^(4)+6(sin theta+cos theta)+4(sin^(6)theta+cos^(6)theta)=13'​

Answer 1

Step-by-step explanation:

=3(sinx−cosx)4+6(sinx+cosx)2+4((sin2x)3+(cos2x)3)

= 3(sin2x+cos2x−2sinxcosx)2+6(sin2x+cos2x+2sinxcosx)+4(sin2x+cos2x)(sin4x+cos4x−sin2xcos2x)

=3(1−sin2x)2+6+6sin2x+4[(sin2x+cos2x)2−3sin2xcos2x]

= 3+3sin22x−6sin2x+6sin2x+6+4−12sin2xcos2x

= 13+3sin22x−3(4sin2xcos2x)

= 13+3sin22x−3(sin22x)

= 13