Question

1) 1.5 sin 2a/cos ( a-1.5П) 2) cos a + cos 2a/sin a - cos a​

Answer 1

1)-3cos(a)

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Answer 2

1)-3cos(a)

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Answer 3

To simplify the given expressions, let's work on each of them step-by-step:

1. Simplify: 1.5 sin(2a) / cos(a - 1.5π)

First, let's use some trigonometric identities: sin(2a) = 2 sin(a) cos(a) cos(a - 1.5π) = cos(a + 1.5π) [cosine is an even function]

Now, the expression becomes: 1.5 * (2 sin(a) cos(a)) / cos(a + 1.5π)

Next, since cos(a + 1.5π) = -sin(a) [using π periodicity of cosine and sine functions],

The expression further simplifies to: 1.5 * (2 sin(a) cos(a)) / (-sin(a))

Now, sin(a) cancels out, and we get: -3 * 2 cos(a) -6 cos(a)

So, the simplified expression is: -6 cos(a)

2. Simplify: (cos(a) + cos(2a)) / (sin(a) - cos(a))

First, let's use some trigonometric identities: cos(2a) = 2 cos^2(a) - 1

Now, the expression becomes: (cos(a) + (2 cos^2(a) - 1)) / (sin(a) - cos(a))

Combine like terms in the numerator: (2 cos^2(a) + cos(a) - 1) / (sin(a) - cos(a))

Now, let's factor the numerator (using factoring by grouping): (2 cos^2(a) - cos(a) + 2 cos(a) - 1) / (sin(a) - cos(a))

Factor out common terms in each group: cos(a) * (2 cos(a) - 1) + 1 * (2 cos(a) - 1) / (sin(a) - cos(a))

Now, factor out (2 cos(a) - 1) from both terms in the numerator: (2 cos(a) - 1) * (cos(a) + 1) / (sin(a) - cos(a))

Now, (cos(a) + 1) can be rewritten as sin(a) using the Pythagorean identity: (2 cos(a) - 1) * sin(a) / (sin(a) - cos(a))

Now, sin(a) cancels out, and we get: 2 cos(a) - 1

So, the simplified expression is: 2 cos(a) - 1

To recap the simplified expressions:

1. 1.5 sin(2a) / cos(a - 1.5π) simplifies to: -6 cos(a)
2. (cos(a) + cos(2a)) / (sin(a) - cos(a)) simplifies to: 2 cos(a) - 1