1.

Use tan addition formula

Results 1 to 15 of 15

- September 3rd 2006, 02:33 AM #1

- Joined
- Jun 2006
- Posts
- 17

## Some question that need help.

1)Without evaluating the angles A and B,find the value of tan(A+2B),

given tanA = 2/3 and tanB =1/3.

2)Simplify (x/y)powerof2/3 . xpowerof3/ypowerof4

3)Find**all**angles between 0deg and 360deg inclusive which satisfy the equation : 3cossquartA - 2cosA - 1 = 0

4)find the value of x when : log(base2)x + log(base8)x = 4/3

BIG thanks!!!

- September 3rd 2006, 02:50 AM #2

- September 3rd 2006, 02:55 AM #3

- September 3rd 2006, 02:59 AM #4

- Joined
- Aug 2006
- From
- Australia
- Posts
- 12

- September 3rd 2006, 03:06 AM #5

- September 3rd 2006, 03:12 AM #6

- September 3rd 2006, 05:08 AM #7

- Joined
- Jun 2006
- Posts
- 17

thx alot...

\log_2{x} + \log_8{x} =\frac{4}{3}

\log_2{x} + \frac{\log_2{x}}{\log_2{8}} =\frac{4}{3}

\log_2{x} + \frac{\log_2{x}}{3} =\frac{4}{3}

\frac{4}{3}\log_2{x} =\frac{4}{3}---->how do you derive from this?care explain further?

\log_2{x} = 1

x=2

3 \cos^2{A} - 2 \cos{A} - 1 = 0

(3 \cos {A} + 1)(\cos{A} - 1)=0

So either \cos{A} = -\frac{1}{3} or \cos{A}=1

yeah i got 109.47deg,289.47deg,70.53 / 0deg,360

am i right ? all anges???

- September 3rd 2006, 05:22 AM #8

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 846

Hello, watsonmath!

Glaysher's solution to #4 is absolutely correct.

I was doing similar problems*before*learning the Base-change Formula.

Because of the 'related' bases (2 and 8), we were taught to*manipulate*the logs.

So here's my baby-talk version.

4) Solve for

Let . .Then:

Take logs (base 2):

. .

Hence, we have: .

The equation becomes: .

- September 3rd 2006, 05:52 AM #9

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- September 3rd 2006, 06:08 AM #10

- Joined
- Jun 2006
- Posts
- 17

- September 3rd 2006, 06:10 AM #11

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- September 3rd 2006, 04:24 PM #12

- Joined
- Jun 2006
- Posts
- 17

- September 3rd 2006, 04:48 PM #13

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- September 4th 2006, 02:04 AM #14

- Joined
- Jun 2006
- Posts
- 17

- September 4th 2006, 04:55 AM #15

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 846

Hello, watsonmath!

Find all angles : .

i got 109.47°, 289.47°, 70.53°, 0°, 360°. .Am i right?

Your second angle is wrong. .(You don't add 180°)

. . And your third angle is also wrong.

If is*negative*, then must be in Quadrant 2 or Quadrant 3.

The expression factors: .

And we have two equations to solve:

. .

. .