Question

Sin2a-2sin4a*cosa +sin6a

Answer 1

$sin2 \alpha +sin6 \alpha =2sin\frac{2 \alpha +6 \alpha }{2}\cdot cos\frac{6 \alpha -2 \alpha }{2}=2sin4 \alpha \cdot cos2 \alpha \\\\\\sin2 \alpha -2sin4 \alpha \cdot cos \alpha +sin6 \alpha =2sin4 \alpha \cdot cos2 \alpha -2sin4 \alpha \cdot cos \alpha =\\\\=2sin4 \alpha (cos2 \alpha -cos \alpha )=2sin4 \alpha \cdot (-2sin\frac{2 \alpha + \alpha }{2}\cdot sin\frac{2 \alpha - \alpha }{2})=\\\\=-4sin4 \alpha \cdot sin\frac{3 \alpha }{2}\cdot sin\frac{ \alpha }{2}$

Answer 2

To simplify the expression Sin(2a) - 2Sin(4a)Cos(a) + Sin(6a), you can use trigonometric identities.

1. Start with the expression:

   Sin(2a) - 2Sin(4a)Cos(a) + Sin(6a)

2. Use the trigonometric identity for the sine of a double angle:

   Sin(2a) = 2Sin(a)Cos(a)

   Replace Sin(2a) with this identity:

   2Sin(a)Cos(a) - 2Sin(4a)Cos(a) + Sin(6a)

3. Now, factor out the common factor of Cos(a):

   2Cos(a)[Sin(a) - Sin(4a)] + Sin(6a)

4. You can simplify further by using the sum-to-product trigonometric identity:

   Sin(p) - Sin(q) = 2Sin[(p - q)/2]Cos[(p + q)/2]

   In this case, let p = 4a and q = a:

   Sin(4a) - Sin(a) = 2Sin[(4a - a)/2]Cos[(4a + a)/2] = 2Sin(3a/2)Cos(5a/2)

   Replace Sin(4a) - Sin(a) with this identity:

   2Cos(a)[2Sin(3a/2)Cos(5a/2)] + Sin(6a)

5. Now, you can simplify the expression further:

   4Cos(a)Sin(3a/2)Cos(5a/2) + Sin(6a)

So, the simplified expression is:

4Cos(a)Sin(3a/2)Cos(5a/2) + Sin(6a)