what is sin^6x + cos^6x
Answers
Answered by
1
Answer:
We have sin
6
x+cos
6
x=(sin
2
x)
3
+(cos
2
x)
3
=(sin
2
x+cos
2
x)
3
−3sin
2
xcos
2
x(sin
2
x+cos
2
x)
=1−3sin
2
xcos
2
x=1−
4
3
.4sin
2
xcos
2
x=1−
4
3
(sin2x)
2
⇒ Maximum value of sin
6
x+cos
6
x is 1−
4
3
×0=1 and minimum values is 1−
4
3
×1=
4
1
Answered by
0
Answer:
is 1
Step-by-step explanation:
The identity used here are: sin^2x+cos^2x=1 ; (a+b)^3=(a^3+b^3+3ab(a+b))
sin^6x+cos^6x=
=(sin^2x+cos^2x)^3–3*(sin^2x)*(cos^2x)*(sin^2x+cos^2x)
= (1)^3–3*(sin^2x)*(cos^2x)*(1)
= 1- 3*(sin^2x)*(cos^2x)
{For the maximum value, The second term,[3*(sin^2x)*(cos^2x)] must be minimum, which is only when x is 0. Therefore the product of sine and cosine term will be equal to 0. }
=1–0
Hence, the maximum value is 1.
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