Appendix B - Trigonometry Review - Exercise Set B - Page A24: 6

$${\text{See explanation}}$$

$$\eqalign{ & 6a) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = 2 \cr & {\text{Adjacent side to }}\theta = 2 \cr & {\text{Hypotenuse }}\theta = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{1}{{\sqrt 2 }},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \sqrt 2 \,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{1}{{\sqrt 2 }},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \sqrt 2 \, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = 1,\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = 1\,\, \cr & \cr & 6b) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = 3 \cr & {\text{Adjacent side to }}\theta = 4 \cr & {\text{Hypotenuse }}\theta = \sqrt {{3^2} + {4^2}} = 5 \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{3}{5},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{5}{3}\,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{4}{5},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{5}{4}\,\, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{3}{4},\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \frac{4}{3} \cr & \cr & 6c) \cr & {\text{From the triangle we can see that:}} \cr & {\text{Opposite side to }}\theta = 1 \cr & {\text{Adjacent side to }}\theta = \sqrt {{4^2} - {1^2}} = \sqrt {15} \cr & {\text{Hypotenuse }}\theta = 4 \cr & {\text{Thus}}{\text{,}} \cr & \cr & {\text{sin}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{1}{4},\,\,\,\,\,{\text{csc}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = 4\,\, \cr & {\text{cos}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{hypotenuse}}}} = \frac{{\sqrt {15} }}{4},\,\,\,\,\,{\text{sec}}\theta = \frac{{{\text{hypotenuse}}}}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{4}{{\sqrt {15} }}\,\, \cr & {\text{tan}}\theta = \frac{{{\text{opposite}}\,{\text{side}}\,\,\theta }}{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }} = \frac{1}{{\sqrt {15} }},\,\,\,\,\,{\text{cot}}\theta = \frac{{{\text{Adjacent}}\,{\text{side}}\,\,\theta }}{{{\text{opposite}}\,{\text{side}}\,\,\theta }} = \sqrt {15} \cr & {\text{See explanation}} \cr} $$