Question

R) 2 sin 3a cos 3a.cos 6a;

Answer 1

Ответ:

$\underbrace{2\cdot sin3a\cdot cos3a}_{sin6a}\cdot \, cos6a=sin6a\cdot cos6a=\dfrac{1}{2}\cdot sin12a$

Answer 2

It seems like you want to simplify the expression:

2sin(3a)cos(3a)cos(6a)

You can use trigonometric identities to simplify this expression. One of the trigonometric identities that will be useful here is the double-angle identity for cosine, which states:

cos(2θ) = 2cos²(θ) - 1

Let's use this identity to simplify your expression step by step:

1. Start with the given expression:

2sin(3a)cos(3a)cos(6a)

2. Use the double-angle identity for cosine with θ = 3a:

cos(6a) = 2cos²(3a) - 1

3. Substitute this result back into the original expression:

2sin(3a)cos(3a)(2cos²(3a) - 1)

4. Now, you have a product of trigonometric functions and a constant (2). You can distribute the 2 into the expression:

4sin(3a)cos(3a)cos²(3a) - 2sin(3a)cos(3a)

5. Now, we can simplify further. Notice that you have a common factor of sin(3a)cos(3a) in both terms:

sin(3a)cos(3a)(4cos²(3a) - 2)

6. Finally, simplify the constant term:

sin(3a)cos(3a)(2(2cos²(3a) - 1))

So, the simplified expression is:

2sin(3a)cos(3a)(2cos²(3a) - 1)