For the sixth question, I got:
$\displaystyle cos a = \frac{-3}{8}$
$\displaystyle cos b = \frac{-5}{4}$
$\displaystyle sin 2a = \frac{7}{8}$
Whenever I use the answer I got, it says it's incorrect, but every time I do it I get that answer.
For the sixth question, I got:
$\displaystyle cos a = \frac{-3}{8}$
$\displaystyle cos b = \frac{-5}{4}$
$\displaystyle sin 2a = \frac{7}{8}$
Whenever I use the answer I got, it says it's incorrect, but every time I do it I get that answer.
Hello, Thomas!
6. Solve the following system.
. . $\displaystyle \begin{array}{ccccccccc}
3\cos a & + & \cos b & + & 5\sin2a & = & 2 & & \\
& & & & \cos a & = & \cos b & + & \sin2a \\
& & 6\cos a & + & 13\sin2a & = & \text{-}7\cos b & & \end{array}$
We have: .$\displaystyle \begin{array}{ccccccc}
3\cos a & + & \cos b & + & 5\sin2a & = & 2 \\
\cos a & - & \cos b & - &\sin2a & = & 0 \\
7\cos a & + & 7\cos b & + & 13\sin2a & = & 0 \end{array}$
And then: .$\displaystyle \begin{vmatrix}
\;3 & 1 & 5 & | & 2 \;\\ \;1 & \text{-}1 & \text{-}1 & | & 0 \;\\ \;7 & 7 & 13 & | & 0\;\end{vmatrix}$
$\displaystyle \begin{array}{c}\text{Switch}\\ R_1\text{ and }R_2 \\ \\ \end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \;\\\;3 & 1 & 5 & | & 2 \;\\\;7 & 7 & 13 & | & 0\;\end{vmatrix}$
$\displaystyle \begin{array}{c} \\ R_2 - 3R_1 \\ R_3 - 7R_1\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0\; \\\;0 & 4 & 8 & | & 2 \;\\\;0 & 14 & 20 & | & 0\;\end{vmatrix}$
. . . $\displaystyle \begin{array}{c} \\ \frac{1}{4}R_2 \\ \frac{1}{2}R_3\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \\0 & 1 & 2 & | & \frac{1}{2} \\0 & 7 & 10 & | & 0\end{vmatrix}$
$\displaystyle \begin{array}{c}R_1+R_2 \\ \\ R_3 - 7R_2\end{array}\;\begin{vmatrix}\;1 & 0 & 1 & | & \frac{1}{2}\;\\
\;0 & 1 & 2 & | & \frac{1}{2}\; \\
\;0 & 0 & \text{-}4 & | & \text{-}\frac{7}{2}\; \end{vmatrix}$
. . . $\displaystyle \begin{array}{c} \\ \\ \text{-}\frac{1}{4}R_3\end{array}\;
\begin{vmatrix}1 & 0 & 1 & | & \frac{1}{2}\\
0 & 1 & 2 & | & \frac{1}{2}\\
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}$
$\displaystyle \begin{array}{c}R_1-R_3 \\ R_2 - 2R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & \text{-}\frac{3}{8} \\
0 & 1 & 0 & | & \text{-}\frac{5}{4} \\
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}$
Therefore: .$\displaystyle \begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} \\ \\\cos b & = & \text{-}\frac{5}{4} \\ \\ \sin2a & = & \frac{7}{8} \end{Bmatrix}$
I had said:
Therefore: .$\displaystyle \begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} & [1] \\ \\\cos b & = & \text{-}\frac{5}{4} & [2]\\ \\ \sin2a & = & \frac{7}{8} & [3]\end{Bmatrix}$
These answers check out perfectly in the three equations.
If there are objections, they should be directed to the problem itself.
Either there are typos in the original statement
. . or the problem was deliberately created to cause confusion.
If [1] is true, then [3] cannot be true.
Also, [2] has no solution in real numbers.
The system is inconsistent and has complex roots.
. . There are no correct answers.
I also agree with your work, I just don't get how we're supposed to answer this question as they are asking for a number (answers are submitted online.) I inputted -3/8 and it said I had the wrong answer.
I've emailed my professor asking about this question.
Thanks for all the help, I'll post back once I have a response from my professor.