# Thread: Find Exact Value of SinB

1. ## Find Exact Value of SinB

Heya!
Ok, here is the exercise:

2. ## Re: Find Exact Value of SinB

$\cos{\beta} = -\cos{\alpha}$

why?

3. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
$\cos{\beta} = -\cos{\alpha}$

why?
Where did u get that?

4. ## Re: Find Exact Value of SinB

Originally Posted by JohnTerry
Where did u get that?
what's the relationship between angles $\alpha$ and $\beta$ ?

5. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
what's the relationship between angles $\alpha$ and $\beta$ ?
But... Isn't smth like:
cosB = 180-cosA [in degrees]

How is cosB = -cosA ?

6. ## Re: Find Exact Value of SinB

Originally Posted by JohnTerry
But... Isn't smth like:
\cos(\beta) = 180-\cos(\alpha) [in degrees]

How is \cos(\beta)= -\cos(\alpha)?
$\beta = \pi - \alpha$

$\cos{\beta} = \cos(\pi - \alpha)$

$\cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}$

$\cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}$

7. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
$\beta = \pi - \alpha$

$\cos{\beta} = \cos(\pi - \alpha)$

$\cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}$

$\cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}$
Then,

cosA= 3/√45

cosB= -(3/√45)

?

8. ## Re: Find Exact Value of SinB

Originally Posted by JohnTerry
Then,

cosA= 3/√45

cosB= -(3/√45)

?
yes ...

you're welcome.

9. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
yes ...

you're welcome.
Thanks, but in the book the answer is -(√5/5)

and -(3/√45) is -(3√45/45)

10. ## Re: Find Exact Value of SinB

Originally Posted by JohnTerry
Thanks, but in the book the answer is -(√5/5)

and -(3/√45) is -(3√45/45)
$-\frac{3}{\sqrt{45}} = -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}$

11. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
$-\frac{3}{\sqrt{45}} = -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}$
I'm sorry, but can you explain please how you passed from $-\frac{3}{\sqrt{9} \cdot \sqrt{5}} to -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$ ?

12. ## Re: Find Exact Value of SinB

Originally Posted by JohnTerry
I'm sorry, but can you explain please how you passed from $-\frac{3}{\sqrt{9} \cdot \sqrt{5}} to -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$ ?
$-\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{3}{3 \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}}$

now rationalize the denominator ...

$-\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

13. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
$-\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{3}{3 \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}}$

now rationalize the denominator ...

$-\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$
Oh! Thanks!

But can you please explain one more thing...
´
How did u get $\cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}$ from $\cos{\beta} = \cos(\pi - \alpha)$

and $\cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}$
from $\cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}$

15. ## Re: Find Exact Value of SinB

Originally Posted by skeeter
Thanks again, but how $\cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}$
= $\cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}$

What does (-1) and (0) mean?

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