Math, asked by prokade606, 3 months ago

sin↑6∅ + cos↑6∅ = 1 - 3sin↑2∅ cos↑2∅

Answers

Answered by Anonymous
7

Answer:

I think , It is an identity to prove.

L

H

S

=

sin

6

x

+

cos

6

x

=

(

sin

2

x

)

3

+

(

cos

2

x

)

3

=

(

sin

2

x

+

cos

2

x

)

3

3

sin

2

x

cos

2

x

(

sin

2

x

+

cos

2

x

)

=

(

1

)

3

3

sin

2

x

cos

2

x

(

1

)

=

1

3

sin

2

x

cos

2

x

=

R

H

S

Proved

Step-by-step explanation:

It is not an equation that can be solved, but rather it is an identity.

Explanation:

Just to help simplify things let

S

=

sin

2

x

Using the fundamental identity

sin

2

A

+

cos

2

A

1

we have;

S

+

cos

2

x

=

1

cos

2

x

=

1

S

So we can then write the initial equation:

sin

6

x

+

cos

6

x

=

1

3

sin

2

x

cos

2

as:

S

3

+

(

1

S

)

3

=

1

3

S

(

1

S

)

S

3

+

1

3

S

+

3

S

2

S

3

=

1

3

S

+

3

S

2

0

=

0

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