Q10 of 80 Page 311

Let us calculate the value of (sin2 α + cos6 α + 3 sin2 α cos2 α).

Given, sin6α + cos6α + 3sin2α cos2α


⇒ sin6α + cos6α + 3sin2α cos2α


= (sin2α)3 + (cos2α)3 + 3sin2α cos2α


= (sin2α + cos2α)(sin4α–sin2αcos2α + cos4α ) + 3sin2α cos2α


= sin4α + cos4α–sin2αcos2α + 3sin2α cos2α


= (sin2α + cos2α)2–2sin2αcos2α–sin2αcos2α + 3sin2α cos2α


= 1


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