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3(sinx-cosx)^4 + 4(sin^6x-cos^6x ) + 6(sinx + cosx )^2 = ????????

3(sinx-cosx)^4 + 4(sin^6x-cos^6x ) + 6(sinx + cosx )^2 = ????????

Grade:12

1 Answers

saket shrivastava
36 Points
9 years ago

open 1st bracket make it into sin2x terms

2nd bracket make a2-b2 open over cube then put expression for a3+b3 and a3-b3 to form cos2x and sin22x terms

3rd bracket again make sin2x

finally we get sin22x and cos2x terms

convert sin to cos to get cubic in cos2x

solve to get 13-8cos6x ans.......................

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