sin↑6∅ + cos↑6∅ = 1 - 3sin↑2∅ cos↑2∅
Answers
Answered by
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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