Question

9) prove that sin6 thita+cos6thita=1-3 sin2 thita cos2 thita​

Answer 1

Step-by-step explanation:

SIN 6 SQUARE THETA + COS 6 SQUARE THETA+3 SIN SQUARE THETA COS SQUARE THETA

LHS=(SIN SQUARE THETA)TO THE POWER 3. +(COS SQUARE THETA)TO THE POWER 3 +3 SIN SQUARE THETA COS SQUARE THETA

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

LHS=(SIN^2THETA + COS^2 THETA)^3. -3 SIN^2 THETA COS^2 THETA ( SIN^2 THETA+ COS SQUARE THETA)^3 + 3 SIN SQUARE THETA COS SQUARE THETA

=LHS=1-3 SIN SQUARE THETA COS SQUARE THETA+3 SIN SQUARE THETA COS SQUARE THETA

=1=RHS

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Answer 2

Answer:

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