sin
Question:

sin6 A + cos6 A + 3 sin2 A cos2 A =

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

(b) 1

We have:

$\sin ^{6} A+\cos ^{6} A+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)$

$=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right) \times 1$

 

$=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)$

$=\left(\sin ^{2} A+\cos ^{2} A\right)^{3}$

 

$=1^{3}=1$

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