Thread: Trigonometry : Stucked @ (b)

1. Trigonometry : Stucked @ (b)

Given that $cos 3x = 4cos^3x -3cosx$,

(a) Find the exact value of $8cos^310 - 6cos10$
(b) solve the equation $8cos^3x - 6cosx = - 1$ for 0< x < 180

Attempted solution

(a) 8cos^310 - 6cos10 = 2(cos3(10))
= 2cos30
= $\sqrt{3}$

(b) since 8cos^3x - 6cosx = 2cos 3x
2cos3x = -1
I cosine inversed $\frac{-1}{2}$ but am not clear on what to do after that.

However, the 3x is giving me some problems on adding and minus...

Note that 0< x <180

2. Originally Posted by Punch
Given that $cos 3x = 4cos^3x -3cosx$,

(a) Find the exact value of $8cos^310 - 6cos10$
(b) solve the equation $8cos^3x - 6cosx = - 1$ for 0< x < 180

Attempted solution

(a) 8cos^310 - 6cos10 = 2(cos3(10))
= 2cos30
= $\sqrt{3}$

(b) since 8cos^3x - 6cosx = 2cos 3x
2cos3x = -1
I cosine inversed $\frac{-1}{2}$ but am not clear on what to do after that.

However, the 3x is giving me some problems on adding and minus...

Note that 0< x <180
I take it the angle is measured in degrees...

$8\cos^3{10} - 6\cos{10} = 2(4\cos^3{10} - 3\cos{10})$

$= 2\cos{(3\cdot 10)}$

$= 2\cos{30}$

$= \frac{2\sqrt{3}}{2}$

$= \sqrt{3}$.

$8\cos^3{x} - 6\cos{x} = -1$

$2\cos{3x} = -1$

$\cos{3x} = -\frac{1}{2}$

Since cosine is negative in the second and third quadrants...

$3x = \{180 - 60, 180 + 60\} + 360n, n \in \mathbf{Z}$

$3x = \{120, 240\} + 360n$

$x = \{40, 80\} + 120n$

In the region $0 < x < 180, x = \{40, 80, 160\}$.

3. Originally Posted by Prove It
I take it the angle is measured in degrees...

$8\cos^3{10} - 6\cos{10} = 2(4\cos^3{10} - 3\cos{10})$

$= 2\cos{(3\cdot 10)}$

$= 2\cos{30}$

$= \frac{2\sqrt{3}}{2}$

$= \sqrt{3}$.

$8\cos^3{x} - 6\cos{x} = -1$

$2\cos{3x} = -1$

$\cos{3x} = -\frac{1}{2}$

Since cosine is negative in the second and third quadrants...

$3x = \{180 - 60, 180 + 60\} + 360n, n \in \mathbf{Z}$

$3x = \{120, 240\} + 360n$

$x = \{40, 80\} + 120n$

In the region $0 < x < 180, x = \{40, 80, 160\}$.
Why do u use 180 - 60?
Isn't $cos^-1\frac {-1}{2}$ = 120?

4. Originally Posted by Punch
Why do u use 180 - 60?
Isn't $cos^-1\frac {-1}{2}$ = 120?
If $\cos{x} = \frac{1}{2}$, the focus angle is $x = 60^{\circ}$.

Since we have $-\frac{1}{2}$, we know that the angle lies in the second or third quadrants.

To work out the angle, for the second quadrant it's $180^\circ - x$, and for the third quadrant it's $180^\circ + x$.

5. hmm and also, how do i know what value to add for the 360n? in $
3x = \{120, 240\} + 360n
$

6. Originally Posted by Prove It
If $\cos{x} = \frac{1}{2}$, the focus angle is $x = 60^{\circ}$.

Since we have $-\frac{1}{2}$, we know that the angle lies in the second or third quadrants.

To work out the angle, for the second quadrant it's $180^\circ - x$, and for the third quadrant it's $180^\circ + x$.
ok so this means that i always have to use to acute angle

7. Originally Posted by Punch
hmm and also, how do i know what value to add for the 360n? in $
3x = \{120, 240\} + 360n
$
There are $360^\circ$ in a circle, and $n$ is an integer representing how many times you have gone around the unit circle.

8. Originally Posted by Prove It
There are $360^\circ$ in a circle, and $n$ is an integer representing how many times you have gone around the unit circle.
From the graphs that I have referred to, for example cos3x=1/2 will mean 3 rounds and that I will have to add 360 twice to each answer i will get.

Thanks you have been a great help although these questions are probably noob to you.

9. Originally Posted by Punch
From the graphs that I have referred to, for example cos3x=1/2 will mean 3 rounds and that I will have to add 360 twice to each answer i will get.

Thanks you have been a great help although these questions are probably noob to you.
No. Read the algebra I posted.

Since you have $3x$, to solve for $x$ you divide EVERYTHING by 3. This includes the 360.