prove that the sin3x= 3sinx-4sincubex
Answers
Answered by
3
Step-by-step explanation:
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
−
−
(
1
)
since we want
sin
3
x
we will expand
(
cos
x
+
i
sin
x
)
3
(
cos
x
+
i
sin
x
)
3
=
cos
3
x
+
3
cos
2
(
i
sin
x
)
+
3
cos
x
(
i
sin
x
)
2
+
(
i
sin
x
)
3
(
cos
x
+
i
sin
x
)
3
=
cos
3
x
−
3
cos
x
sin
2
x
+
i
(
3
cos
2
x
sin
x
−
sin
3
x
)
from
(
1
)
cos
3
x
+
i
sin
3
x
=
cos
3
x
−
3
cos
x
sin
2
x
+
i
(
3
cos
2
x
sin
x
−
sin
3
x
)
equating imaginary parts
sin
3
x
=
3
cos
2
x
sin
x
−
sin
3
x
but
cos
2
x
=
1
−
sin
2
x
sin
3
x
=
3
(
1
−
sin
2
x
)
sin
x
−
sin
3
x
sin
3
x
=
3
sin
x
−
3
sin
2
x
sin
x
−
sin
3
x
sin
3
x
=
3
sin
x
−
4
sin
3
x
as required
I hope it will help for this sum
Answered by
1
Answer:
sorry l don't know
Step-by-step explanation:
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