# Special Triangles Question

• June 1st 2011, 05:51 AM
starshine84
Special Triangles Question
I am looking for help with this question from my grade 11 advanced functions course:

csc θ = -17/15 where 270° ≤ θ ≤ 360° and
cot β = - 3/4 where 90° ≤ β ≤ 180°.

Find the exact value of sin(θ + β). Show all the work.
• June 1st 2011, 05:54 AM
Prove It
Better yet, why don't you show all of YOUR work, i.e. what you have tried and where exactly you are stuck... Then you might get the help you need...
• June 1st 2011, 06:03 AM
starshine84
This what I have done:

cscθ = -17/15 where 270° ≤ θ ≤ 360°

sinθ = 1/cscθ
sinθ = 1/(-17/15)
sinθ = 1/1 / (-17/15)
sinθ = 1/1 * (-15/17)
sinθ = -15/17
θ = sin^-1 (-15/17)

** I don't know what to do from here because if I solve for θ completely I won't have an EXACT value **

cot β = - 3/4 where 90° ≤ β ≤ 180°

tanβ = 1/cotβ
tanβ = 1/(-3/4)
tanβ = 1/1 / (-3/4)
tanβ = 1/1 * (-4/3)
tanβ = -4/3
β = tan^-1 (-4/3)

** I don't know what to do from here because if I solve for β completely I won't have an EXACT value **

From this point I'm not sure how to find an exact value for sin(θ + β) because I do not have values for θ or β
• June 1st 2011, 06:19 AM
Prove It
For starters, you should know that $\displaystyle \sin{(\theta \pm \beta)} \equiv \sin{(\theta)}\cos{(\beta)} \pm \cos{(\theta)}\sin{(\beta)}$.

You don't need to solve for $\displaystyle \theta$ or $\displaystyle \beta$, you only need to solve for $\displaystyle \sin{(\theta)}, \cos{(\theta)}, \sin{(\beta)}$ and $\displaystyle \cos{(\beta)}$. You will need to make use of the Pythagorean Identity.
• June 1st 2011, 07:46 AM
starshine84
I knew about the sum and difference formulas and that I would have to use the sum formula for sin in order to solve the final step, but I'm not sure what you mean by pythagorean identities. I'm not sure how to solve for sin(\theta), cos(\theta), sin(\beta), or cos(\beta).
• June 1st 2011, 08:12 AM
Plato
Quote:

Originally Posted by starshine84
This what I have done:
cscθ = -17/15 where 270° ≤ θ ≤ 360°
cot β = - 3/4 where 90° ≤ β ≤ 180°
** I don't know what to do from here because if I solve for β completely I won't have an EXACT value

From that given you know that $\theta\in IV$.
So $\sin \left( \theta \right) = \frac{{ - 15}}{{17}}\;\& \,\cos \left( \theta \right) = \sqrt {1 - \left( {\frac{15}{17}} \right)^2 }=\frac{8}{{17}}$

Moreover, $\beta\in II$ so $\sin \left( \beta \right) = \frac{4}{5}\;\& \,\cos \left( \beta \right) = \frac{{ - 3}}{5}$.
• June 1st 2011, 08:41 AM
starshine84
So would it then be...

sin(\theta+\beta) = sin\theta cos\beta + cos\theta sin\beta
sin(\theta+\beta) = (-15/17)(-3/4) + (8/17)(4/5)
sin(\theta+\beta) = (45/68) + (32/85)
sin(\theta+\beta) = 225/340 + 128/340
sin(\theta+\beta) = 353/340

??
• June 1st 2011, 02:58 PM
Prove It
Looks good to me :)
• June 2nd 2011, 04:05 AM
starshine84
That's awesome, thank you so much for all of your help!