Simplify the expression:
\[
\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}
\]
Answer (3)
2 co s 3 α − cos α s in α − 2 s i n 3 α = cos α ( 2 co s 2 α − 1 ) s in α ( 1 − 2 s i n 2 α ) = cos α ( co s 2 α + co s 2 α − 1 ) s in α ( 1 − s i n 2 α − s i n 2 α ) = cos α ( co s 2 α − s i n 2 α ) s in α ( co s 2 α − s i n 2 α ) = cos α s in α = t an α s i n 2 α + co s 2 α = 1 → co s 2 α = 1 − s i n 2 α s i n 2 α + co s 2 α = 1 → s i n 2 α = 1 − co s 2 α → − s i n 2 α = co s 2 α − 1
To simplify the given expression Sin A - 2 sin cube A / (2 cos cube A - cos A), we need to use the required identities. The simplified expression is: tanA
To simplify the given expression Sin A - 2 sin cube A / (2 cos cube A - cos A), we can use the identities: sin³ A = (sin A)² sin A and cos³ A = (cos A)² cos A.
Applying these identities, we get: Sin A - 2 sin cube A = sin A - 2(sin A)² sin A = sin A(1 - 2sin² A)
Simplify: 2 cos cube A - cos A = 2(cos A)² cos A - cos A = cos A(2 cos² A - 1)
Now, we can rewrite the expression as: (sin A(1 - 2sin² A)) / (cos A(2 cos² A - 1))
Let's manipulate the numerator and denominator:
cos A ( 1 − 2 sin 2 A ) sin A ( 1 − 2 sin 2 A )
Now, notice that we can factor out 1 − 2 s i n 2 A from both the numerator and denominator:
SinA / CosA
This is equal to tanA.
The expression 2 c o s 3 A − c o s A s i n A − 2 s i n 3 A simplifies to tan A by factoring both the numerator and denominator and canceling the common terms. The use of double angle identities aids in the simplification process. Thus, the final simplified form is tan A .
;
