sum and difference identities

[trigfun]

Hi,
I'm having trouble with this problem. I know the right answer but I can't figure out how to get there.

"Find the exact value of each of the following under the given conditions:

sin(alpha) = [MATH]5/13[/MATH] (0 < alpha < pi/2)
cos(beta) = [MATH](2*sqrt53)/53[/MATH] ( -pi/2 < beta < 0)

sin(alpha+beta) = [MATH](-74*sqrt53)/689[/MATH]cos(alpha + beta) = [MATH](59*sqrt53)/689[/MATH]sin(alpha-beta) = [MATH](94*sqrt53)/689[/MATH]tan(alpha - beta)= [MATH]-94/11[/MATH]
I'm using these formulas:
[MATH]sin(a+b) = sina*cosb+cosa*sinb[/MATH][MATH]cos(a+b) = cosa*cosb-sina*sinb[/MATH][MATH]sin(a-b) = sina*cosb-cosa*sinb[/MATH][MATH]tan(a+b) = (tana + tanb)/1+tana*tanb[/MATH]
Can anyone show me how to get to these answers using these formulas?

Thank you!

 

[Romsek]

I'm assuming that you are given

[MATH]\sin(\alpha) = \dfrac{5}{13},~0 \leq \alpha < \dfrac \pi 2\\ \cos(\beta) = \dfrac{2\sqrt{53}}{53},~-\dfrac \pi 2 < \beta \leq 0[/MATH]
and you are to calculate for example
[MATH]\sin(\alpha + \beta)[/MATH]
You're going to need
[MATH]\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}\\ \sin(\alpha) = \sqrt{1 - \cos^2(\beta)} = \sqrt{1 - \dfrac{4\cdot 53}{53^2}} = \sqrt{1 - \dfrac{4}{53}} = \sqrt{\dfrac{49}{53}}=\dfrac{7\sqrt{53}}{53}[/MATH]
now just plug and chug

[MATH]\sin(\alpha+\beta) = \dfrac{5}{13}\dfrac{2\sqrt{53}}{53}+\dfrac{12}{13}\dfrac{7\sqrt{53}}{53} = ?[/MATH]
The others are similar.

 

[trigfun]

Thank you! I can see where I was making mistakes. I appreciate the help!

 

[trigfun]

Wait, now I have more questions. For sin(alpha+beta) above, I keep getting (94*sqrt(53))/689.
The answer that I'm being told is correct is (-74*sqrt(53))/689.

The only way I've been able to get the right answer is if I subtract one side from the other instead of add - but that's not what the identity says to do.

Am I missing something obvious?

Thank you!

(sorry for the lack of proper symbols, I haven't figured out how to do that part yet)

 

[Romsek]

oh I see I led you astray.

as [MATH]\beta[/MATH] is in the fourth quadrant [MATH]\sin(\beta) = -\dfrac{7\sqrt{53}}{53}[/MATH]
sorry about that chief

 

[Steven G]

I'm assuming that you are given

[MATH]\sin(\alpha) = \dfrac{5}{13},~0 \leq \alpha < \dfrac \pi 2\\ \cos(\beta) = \dfrac{2\sqrt{53}}{53},~-\dfrac \pi 2 < \beta \leq 0[/MATH]
and you are to calculate for example
[MATH]\sin(\alpha + \beta)[/MATH]
You're going to need
[MATH]\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}\\ \sin(\alpha) = \sqrt{1 - \cos^2(\beta)} = \sqrt{1 - \dfrac{4\cdot 53}{53^2}} = \sqrt{1 - \dfrac{4}{53}} = \sqrt{\dfrac{49}{53}}=\dfrac{7\sqrt{53}}{53}[/MATH]
now just plug and chug

[MATH]\sin(\alpha+\beta) = \dfrac{5}{13}\dfrac{2\sqrt{53}}{53}+\dfrac{12}{13}\dfrac{7\sqrt{53}}{53} = ?[/MATH]
The others are similar.

To OP,
In the 8th line Romsek meant to wrote \(\displaystyle \sin(\beta)=...\)