Question

Sin 10a – 2 sin 6a cos 6a .1-2 sin²6a - cos 10 a​

Answer 1

Ответ:

в чем заключается вопрос? Если надо упростить то вот:

Answer 2

It looks like you've provided an expression involving trigonometric functions. Let's simplify it step by step:

Given expression: sin(10a) - 2sin(6a)cos(6a) - (1 - 2sin²(6a) - cos(10a))

Step 1: Use trigonometric identities

1. sin(10a) can be rewritten as sin(6a + 4a).
2. Use the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ).
3. Use the double angle formula for cosine: cos(2θ) = cos²(θ) - sin²(θ).

Step 2: Apply the trigonometric identities sin(6a + 4a) - 2(2sin(6a)cos(6a)) - (1 - 2sin²(6a) - (cos²(5a) - sin²(5a)))

Step 3: Simplify further sin(6a)cos(4a) + 2sin(6a)cos(6a) - (1 - 2sin²(6a) - cos²(5a) + sin²(5a))

Step 4: Expand the expressions involving angles (sin(6a)cos(4a)) + (2sin(6a)cos(6a)) - 1 + 2sin²(6a) + cos²(5a) - sin²(5a)

Step 5: Use additional trigonometric identities

1. cos(4a) = cos²(2a) - sin²(2a).
2. sin²(5a) + cos²(5a) = 1 (Pythagorean identity).

Step 6: Substitute the identities (sin(6a)(cos²(2a) - sin²(2a))) + (2sin(6a)cos(6a)) - 1 + 2sin²(6a) + 1 - sin²(5a))

Step 7: Distribute and simplify sin(6a)cos²(2a) - sin(6a)sin²(2a) + 2sin(6a)cos(6a) + 2sin²(6a) - sin²(5a)

Step 8: Use trigonometric identity sin(2θ) = 2sin(θ)cos(θ).

Step 9: Substitute the identity sin(6a)cos²(2a) - sin(6a)sin²(2a) + 2sin(6a)cos(6a) + 2sin²(6a) - sin²(5a) = sin(6a)[2cos(2a)sin(2a)] + 2sin(6a)cos(6a) + 2sin²(6a) - sin²(5a)

Step 10: Simplify the first term using the double angle formula = sin(6a)sin(4a) + 2sin(6a)cos(6a) + 2sin²(6a) - sin²(5a)

Step 11: Use the double angle formula for sine sin(6a)sin(4a) = (1/2)[cos(2a) - cos(10a)]

Step 12: Substitute the result back into the expression (1/2)[cos(2a) - cos(10a)] + 2sin(6a)cos(6a) + 2sin²(6a) - sin²(5a)

This is the simplified form of the expression you provided. If you have specific values for 'a' that you want to substitute, you can plug them in to get a numerical result.