Prove the following identities:
Answers
cos3A=4cos
3
A−3cosA
and
sin3A=3sinA−4sin
3
A
Using these in LHS,
∴ The L.H.S =cos
3
A(4cos
3
A−3cosA)+sin
3
A(3sinA−4sin
3
A)
⇒4cos
6
A−3cos
4
A+3sin
4
A−4sin
6
A
⇒4(cos
6
A−sin
6
A)−3(cos
4
A−sin
4
A)
⇒4{(cos
2
A)
3
−(sin
2
A)
3
}−3{(cos
2
A)
2
−(sin
2
A)
2
}
⇒4{cos
2
A−sin
2
A}{(cos
2
A)
2
+cos
2
Asin
2
A+(sin
2
A)
2
}−3(cos
2
A−sin
2
A)(cos
2
A+sin
2
A)
⇒(cos
2
A−sin
2
A)[4{(cos
2
A)
2
+cos
2
Asin
2
A+(sin
2
A)
2
}−3.1]
⇒(cos
2
A−sin
2
A)[4cos
2
A+4cos
2
Asin
2
A+4sin
4
A−3(cos
2
A+sin
2
A)
2
]
⇒(cos
2
A−sin
2
A)[4cos
4
A+4cos
2
Asin
2
A+4sin
4
A−3(cos
4
A+2cos
2
Asin
2
A+sin
4
A)]
⇒(cos
2
A−sin
2
A)[cos
4
A−2cos
2
Asin
2
A+sin
4
A]
⇒(cos
2
A−sin
2
A)(cos
2
A−sin
2
A)
2
,
⇒(cos
2
A−sin
2
A)
3
⇒(cos2A)
3
⇒cos
3
2A
=RHS Hence proved