Derive the formula for cot 3A
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Answer:
cot3A=
sin3A
cos3A
⟶(1)
We know that
cos3A=4cos
3
A−3cosA⟶(2)
sin3A=3sinA−4sin
3
A⟶(3)
Substituting (2) and (3) in (1) we get
cot3A=
3sinA−4sin
3
A
4cos
3
A−3cosA
=
sinA(3−4sin
2
A)
cosA(4cos
2
A−3)
=cotA(
3−4sin
2
A
4cos
2
A−3
)
=cotA(
3(sin
2
A+cos
2
A)−4sin
2
A
4cos
2
A−3×(sin
2
A+cos
2
A)
)(∵sin
2
A+cos
2
A=1)
=cotA(
3cos
2
A+3sin
2
A−4sin
2
A
4cos
2
A−3cos
2
A−3sin
2
A
)
=cotA(
3cos
2
A−sin
2
A
cos
2
A−3sin
2
A
)
=cotA×
sin
2
A
sin
2
A
(
3cos
2
A/sin
2
A−1
cos
2
A/sin
2
A−3
)
=cotA(
3cot
2
A−1
cot
2
A−3
)
=
3cot
2
A−1
cot
3
A−3cotA
=
1−3cot
2
A
3cotA−cot
3
A
Hence proved.
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