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# Find the Exact Value

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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

May 10, 2019
edited by AdamTaurus  May 10, 2019
edited by AdamTaurus  May 10, 2019

#1
+8853
+4

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.

$$\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}$$

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May 10, 2019

#1
+8853
+4

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.

$$\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}$$

hectictar May 10, 2019