**Q: Find the exact value of sin(2a) and cos(2a) using the double angle formulas, given sin(a) = 5/9 , π/2 < a < π**

A: We are told to use the double angle identity. First I will solve for sin(2a).

The double angle identity says:

sin(2a) = 2*sin(a)*cos*(a)

OK, well I know that sin(a) = 5/9, so I need to know what cos(a) is. To solve for this, draw a right tringle and lable one of the angles “a”. Since sin is the oppostite side / hypotenuse, I know that one side of the triangle is 5 and the hypotenuse is 9. Use Pythagorean Theorem to solve for the missing side, which I will call “b”.

a² + b² = c²

5² + b² = 9²

25 + b² = 81

b² = 66

b = √(56) = √(4)*√(14) = 2√(14)

So, cos(a) = adjacent/hypotenuse = 2√(14) / 9

HOWEVER… Remember that: π/2 < a < π. This tells us that “a” is in the second quadrant. All cosine values in the second quadrant should be negative, therefore: **cos(a) = – 2√(14) / 9**

OK. Back to the problem:

sin(2a) = 2*sin(a)*cos(a)

Now substitute in our numbers:

sin(2a) = 2*(5/9)*(-2√(14) / 9)

Multiply the fractions across the top and across the bottom to get:

**sin(2a) = -20√(14) / 81**

Now to solve cos(2a).

There are many double angle identities for cosine, so I am just going to pick one:

cos(2a) = 2cos²a – 1

Plug in our numbers:

cos(2a) = 2(-2√(14) / 9)² – 1

Square everything in the parentheses:

cos(2a) = 2(4*14 / 81) – 1

cos(2a) = 112 / 81 – 1

**cos(2a) = 31/81**