# Trigonometry Problems

• Oct 3rd 2012, 07:36 PM
Vold
Trigonometry Problems
I've become stumped on these last two problems and would appreciate any and all help.

1)Find all the solutions in degrees to 21sin2(theta)+10sin(theta)=-1. Round your answers to one decimal place.

2) Let pi < alpha < 3pi/2 ; pi/2 < beta < pi ; cos(alpha) = -4/9 ; sin(beta)= 7/12. Find the exact value for each of the following:

cos(alpha-beta)
sin(alpha+beta)
cos(alpha/2)
sin(2*alpha)

Thank you.
• Oct 3rd 2012, 07:49 PM
MarkFL
Re: Trigonometry Problems
1.) You have a quadratic in $\displaystyle \sin(\theta)$ which factors nicely.

2.) You will need angle-sum identities for sine and cosine, half-angle identity for cosine and double-angle identity for sine. But first, you will need the Pythagorean identity and the information about the quadrants in which the given angles reside to determine $\displaystyle \sin(\alpha)$ and $\displaystyle \cos(\beta)$.

Let's see what you come up with...
• Oct 3rd 2012, 07:57 PM
Vold
Re: Trigonometry Problems
I've figured out the first question with the help of factoring the quadratic. My confusion lies in the second problem and how to use the Pythagorean identity to determine sin(a) and cos(B). The problem will be easy enough then to start plugging into the sum and difference formulas, etc.
• Oct 3rd 2012, 08:04 PM
MarkFL
Re: Trigonometry Problems
We know:

$\displaystyle \sin^2(\alpha)=1-\cos^2(\alpha)=1-\left(-\frac{4}{9} \right)^2=\frac{65}{81}$

Since $\displaystyle \alpha$ is in the third quadrant, we know $\displaystyle \sin(\alpha)$ is negative. So we find:

$\displaystyle \sin(\alpha)=-\frac{\sqrt{65}}{9}$

See if you can use the same reasoning to find $\displaystyle \cos(\beta)$.
• Oct 3rd 2012, 08:21 PM
Vold
Re: Trigonometry Problems
Well, I got Cos(B)= Sqrt95/12, so assuming that's correct I went ahead and plugged everything into the formulas for the rest of the question. Appreciate your help, thank you.
• Oct 3rd 2012, 08:36 PM
MarkFL
Re: Trigonometry Problems
$\displaystyle \beta$ is in the second quadrant, so $\displaystyle \cos(\beta)$ will be negative. (Nod)
• Oct 3rd 2012, 08:39 PM
Vold
Re: Trigonometry Problems
... close enough. Hahah thank you.