(cos^6 theta +sin^6theta)+3sin^2 theta cos^2theta =?
Answers
Answered by
1
Answer:
sorry can't understand your question properly
Answered by
0
Answer:
sin
6
θ+cos
6
θ+3sin
2
θcos
2
θ
⇒LHS=(sin
2
θ)
3
+(cos
2
θ)
3
+3sin
2
θcos
2
θ
Using, [a
3
+b
3
=(a+b)
3
−3ab(a+b)]
⇒LHS=(sin
2
θ+cos
2
θ)
3
−3sin
2
θcos
2
θ(sin
2
θ+cos
2
θ)
3
+3sin
2
θcos
2
θ
⇒LHS=1−3sin
2
θcos
2
θ+3sin
2
θcos
2
θ=1=RHS
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