Math, asked by sapnabansalkhirwal, 2 months ago

what is sin^6x + cos^6x​

Answers

Answered by farhaanaarif84
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 reenusingh1981
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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