Could someone please post a solution to these 4 questions:

http://img413.imageshack.us/img413/6047/trigtestyq3.png

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- February 6th 2007, 07:47 PMSportfreundeKeaneKentProblems From Trig Test
Could someone please post a solution to these 4 questions:

http://img413.imageshack.us/img413/6047/trigtestyq3.png - February 7th 2007, 01:07 PMearboth

Hello,

to #3.a)

(For some reason the Latex doesn't work properly anymore. I'll repeat the equations in plain text)

cos(2x) = 0 or cos(2x) = -1 or cos(2x) = 1

This character π means pi

From the 1rst equ.:

2x = -π/2 --> x = -π/4 or

2x = -3π/2 ---> x = -3π/4

from the 2nd equ.:

2x = -π ---> x = -π/2

from the 3rd equ.:

2x = 0 ---> x = 0 or

2x = -2π ---> x = -π

EB - February 7th 2007, 01:58 PMSoroban
Hello, SportfreundeKeaneKent!

Here's the very first one . . .

Quote:

1. If cos(x) = -12/13, where π__<__x__<__3π/2

then find the exact value of sin(2x)

We'll use the identity: .sin(2x) .= .2·sin(x)·cos(x)

We know that: cos(x) = -12/13 . . . we need sin(x).

Angle x is in Quadrant 3.

Since cos(x) = -12/13 = adj/hyp,

. . we use Pythagorus and find that: opp = -5

Hence: .sin(x) = -5/13

Therefore: .sin(2x) .= .2(-5/13)(-12/13) .= .**120/169**

- February 7th 2007, 03:21 PMSoroban
Hello again, SportfreundeKeaneKent!

Here's #4(b) . . .

Quote:

Prove:

. . . . . . . . . . . . . .tan(x) - tan(y)

. . tan(x - y) . = . --------------------

. . . . . . . . . . . . .1 + tan(x)·tan(y)

. . sin(x - y) .= .sin(x)·cos(y) - sin(y)·cos(x)

. . cos(x - y) .= .cos(x)·cos(y) + sin(x)·sin(y)

. . . . . . . . . . . sin(x - y) . . . . . sin(x)·cos(y) - sin(y)·cos(x)

tan(x - y) . = . ------------ . = . -----------------------------------

. . . . . . . . . . . cos(x - y) . . . . .cos(x)·cos(y) + sin(x)·sin(y)

Divide top and bottom by cos(x)·cos(y):

. . .sin(x)·cos(y) . .sin(y)·cos(x) . . . . . . sin(x) . sin(y)

. . .--------------- - --------------- . . . . . . ------- - -------

. . .cos(x)·cos(y) . cos(x)·cos(y) . . . . . . cos(x) . cos(y)

. . ------------------------------------ . = . ---------------------

. . .cos(x)·cos(y) . .sin(x)·sin(y) . . . . . . . . sin(x) sin(y)

. . .--------------- + --------------- . . . . .1 + -------·--------

. . .cos(x)·cos(y) . .cos(x)·cos(y) . . . . . . . .cos(x) cos(y)

. . . . . . . tan(x) - tan(y)

. . . = . ---------------------

. . . . . . 1 + tan(x)·tan(y)

- February 8th 2007, 12:45 AMearboth
- February 8th 2007, 01:04 AMearboth
Hello, Sportsfreund,

#2

Use the property:

As Soroban has demonstrated use the Pythagoran theorem:

. Because x is in the 4. quadrant the sine-value must be negative.

. Because w is in the 2. quadrant the cosine-value must be negative.

Now you know all values necessary:

EB - February 8th 2007, 06:26 AMSoroban
Hello, SportfreundeKeaneKent!

There's something wrong with 4(a).

. . Is there a typo?

Quote:

The right side is always equal to 1.

. . So it's a*silly*way to write a problem.

Besides, the statement is**not**an identity.

Try x = π/3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._

The left side is: .sec(π/3)[1 - sin(π/3)] .= .2(1 - √3/2) . ≠ . 1