Algebra 2/Trigonometry Help. IF you can.?

Question

Hey there, I am enrolled in Calc/Phys...and our teacher gave us some trig/algebra problems to complete over the summer because we are having a big test 1st day of school. I was just stuck on a few, and after some thoughts I clearly forgot how to do them. I tried and looked up different ways of solving the types of problems, but some weren't there. Can you guys help me solve the following 3 problems please?

1. Find Sin2A if SinA=1/4 and 0<A<90

Answer: (Root15)/8

2. If Cos2A=1/3, and 0<2A<180, find cos A

Answer: Root(2/3)

3. Write the Given expression in Algebraic form:

Tan(arccos(x/3))

Answer: Root(9-x^2)/x

Much help is appreciated fellow yahoo members! Thanks!

Answer 1

[All 3 answered now.]

1.

sin2A = 2sinAcosA

If sinA = 1/4, then 1 is one side of a triangle and 4 is the hypotenuse, therefore you can find cosA:

1² + x² = 4²

x² = 15

x = √(15)

cosA = √(15)/4

sin2A = 2(1/4)(√(15)/4)

sin2A = √(15)/8

2.

cos2A = 1 - 2sin²A

1/3 = 1 - 2sin²A

-2/3 = - 2sin²A

1/3 = sin²A

Using identity sin²A + cos²A = 1:

1/3 + cos²A = 1

cos²A = 2/3

cosA = √(2/3)

3.

This is tricky.  There is an identity that goes along with that, however you probably don&#039;t know it yet.

tan[arccos(x/3)]

= tan [theta/some angle]

To make this easier, first replace x/3 with A.  What you know is that A/1 is cosine.  So, draw a  right triangle with angle theta, hypotenuse 1, and the adjacent side to theta put A.  Now, solve for the opposite side, named c:

A² + c² = 1²

c² = 1 - A²

c = √(1 - A²)

Now you can solve for tan[theta].  You know that tan is opposite over adjacent sides, so oppose is √(1 - A²), and adjacent is A:

tan [theta] = [√(1 - A²)]/A   &lt;---(This is the identity for tan[arccosx] by the way.  Your teacher might give you brownie points for finding it.  :P )

Now you can put back in x/3:

tan [arccos(x/3)] = √(1 - (x/3)²)]/(x/3)

= 3√(1 - (x/3)²)]/x

Just factor in the 3:

= √(9(1 - x²/9)]/x

= √(9 - x²)]/x