Question

Cos3a/2sina + 3sina/2cosa

Answer 1

$\frac{cos(3a)}{2sin(a)}+ \frac{3sin(a)}{2cos(a)} = \frac{cos(3a)2cos(a)+3sin(a)2sin(a)}{4sin(a)cos(a)} = \frac{2(cos(3a)cos(a)+3sin^2(a)}{4sin(a)cos(a)} =\\= \frac{cos(3a)cos(a)+3sin^2(a)}{sin(2a)}$

Answer 2

We can start by using the trigonometric identity cos(3a) = 4cos^3(a) - 3cos(a) to simplify the first term:

cos(3a) / (2sin(a)cos(a)) = (4cos^3(a) - 3cos(a)) / (2sin(a)cos(a))

Next, we can simplify the denominator by factoring out a 2 from the denominator:

(4cos^3(a) - 3cos(a)) / (sin(a)cos(a)) = (2cos(a))(2cos^2(a) - 3) / (sin(a)cos(a))

We can simplify further by canceling out the common factor of cos(a) from the numerator and denominator:

(2cos^2(a) - 3) / sin(a)

Now, let's simplify the second term:

3sin(a) / (2cos(a))

We can simplify the denominator by factoring out a 2:

3sin(a) / (2cos(a)) = (3/2)(sin(a)/cos(a))

Using the identity tan(a) = sin(a)/cos(a), we can rewrite this as:

(3/2)tan(a)

So, the original expression can be simplified to:

(2cos^2(a) - 3) / sin(a) + (3/2)tan(a)

This is the simplified form of the original expression.