+895 Sorry, idk why the formatting won't show on the domains when I do Latex.
Find the exact value of each of the following under the given conditions:
1)\(sin(a+B)\) 2)\(cos(a+B)\) 3)\(sin(a-B)\) 4)\(tan(a-B)\)
\(sin(a)=\frac{5}{13}\)
-3pi/2 < a < -pi
\(tan(B)=-\sqrt3\)
pi/2 < B < pi
+9479 To use the sum and difference formulas, we need to find cos(a), cos(B), sin(a), sin(B), tan(a) and tan(B).
Using the information given, we can find these values with the Pythagorean theorem.
\(\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}\)
+9479 To use the sum and difference formulas, we need to find cos(a), cos(B), sin(a), sin(B), tan(a) and tan(B).
Using the information given, we can find these values with the Pythagorean theorem.
\(\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}\)
