Question

How do you use double angle formulas to calculate cos 2x and sin 2x without finding x if `cos x = 3/5` and x is in the first quadrant?

Answer 1

`cos(2x)=-7/25`
`sin(2x)=24/25`

Explanation:

The double-angle formulas state that:
`cos(2x)=cos^2(x)-sin^2(x)`

`sin(2x)=2sin(x)cos(x)`

Now, we are given that `cos(x)=3/5`. Then, knowing that `sin^2(x)+cos^2(x)=1` and that `x` is in the first quadrant (and thus `sin(x)` and `cos(x)` are both positive), we can find that `sin(x)=sqrt(1-cos^2(x))=4/5`.

Thus, using the identities mentioned above,
`cos(2x)=cos^2(x)-sin^2(x)=(3/5)^2-(4/5)^2=-7/25`

`sin(2x)=2sin(x)cos(x)=2*4/5*3/5=24/25`

Answer 2

`cos2x = -7/25`
`sin2x = 24/25`

Explanation:

We have `cosx=3/5` and `x` is acute.

Using `sin^2x + cos^2x -= 1`, we have:

`sin^2x + (3/5)^2 = 1`
`=> sin^2x = 1-9/25`
`:. \ sin^2x = 16/25`
`:. \ \ sinx = +-sqrt(16/25)`

And knowing that `x` is acute we take the positive solution.

`=> sinx = 4/5`

Next we use the sine and cosine double angle formula, so that:

`cos2x -= cos^2x - sin^2 x`

`\ \ \ \ \ \ \ \ \ = (3/5)^2 - (4/5)^2`

`\ \ \ \ \ \ \ \ \ = 9/25 - 16/25`

`\ \ \ \ \ \ \ \ \ = -7/25`

And:

`sin2x -= 2sinxcosx`

`\ \ \ \ \ \ \ \ \ = 2(3/5)(4/5)`

`\ \ \ \ \ \ \ \ \ = 24/25`