Find the value of the trigonometric expressions
- Thread starter tanasaur
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Do you know the formula for the sum of the cosine of two angles, i.e. Cos (a + b)?
For the trig values given, draw a triangle representing these ratios in the proper quadrant in order to substitute the proper trig values needed into the Cos (a + b) formula.
P.S. You will have to also know the identity for Cos(2x) and Sin (2x) as well
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Hello, tanasaur!
\(\displaystyle \tan\alpha\text{ is negative in Quadrants 2 and 4.}\)
\(\displaystyle \sin\alpha\text{ is positive in Quadrants 1 and 2.}\)
. . \(\displaystyle \text{Hence: }\,\alpha\text{ is in Quadrant 2.}\)
\(\displaystyle \tan\alpha \:=\:\dfrac{+1}{-2} \:=\:\dfrac{opp}{adj} \)
\(\displaystyle \alpha\text{ is in Quadrant 2 with: }\,opp = 1,\:adj = -2\)
. . \(\displaystyle \text{Hence: }\,hyp = \sqrt{5}\)
\(\displaystyle \sin\alpha \,=\,\dfrac{1}{\sqrt{5}},\;\cos\alpha \,=\,-\dfrac{2}{\sqrt{5}} \;\;[1]\)
\(\displaystyle \text{Since }\,\frac{3\pi}{2} < \beta < 2\pi,\;\beta\text{ is in Quadrant 4.}\)
\(\displaystyle \csc\beta \,=\,\dfrac{+6}{-5} \,=\,\dfrac{hyp}{opp}\)
\(\displaystyle \beta\text{ is in Quadrant 4 with: }\,opp = -5,\;hyp = 6\)
. . \(\displaystyle \text{Hence: }\,adj = \sqrt{11}\)
\(\displaystyle \sin\beta \,=\,-\dfrac{5}{6},\;\cos\beta \,=\,\dfrac{\sqrt{11}}{6}\;\;[2] \)
\(\displaystyle \cos(2\alpha + \beta) \;=\;\cos(2\alpha)\cos(\beta) - \sin(2\alpha)\sin(\beta) \)
. . . . . . . . . .\(\displaystyle =\;(\cos^2\!\alpha - \sin^2\!\alpha)\cos\beta - (2\sin\alpha\cos\alpha)\sin\beta\)
. . . . . . . . . .\(\displaystyle =\;\bigg[\left(\dfrac{-2}{\sqrt{5}}\right)^2 - \left(\dfrac{1}{\sqrt{5}}\right)^2\bigg]\dfrac{\sqrt{11}}{6} - \bigg[2\left(\dfrac{1}{\sqrt{5}}\right)\left(\dfrac{-2}{\sqrt{5}}\right)\bigg]\left(-\dfrac{5}{6}\right) \)
. . . . . . . . . .\(\displaystyle =\;\left(\dfrac{4}{5} - \dfrac{1}{5}\right)\left(\dfrac{\sqrt{11}}{6}\right) - \left(-\dfrac{4}{5}\right)\left(-\dfrac{5}{6}\right) \)
. . . . . . . . . .\(\displaystyle =\;\dfrac{\sqrt{11}}{10} - \dfrac{2}{3}\)
. . . . . . . . . .\(\displaystyle =\;\dfrac{3\sqrt{11} - 20}{30}\)
\(\displaystyle \tan\alpha\text{ is negative in Quadrants 2 and 4.}\)
\(\displaystyle \sin\alpha\text{ is positive in Quadrants 1 and 2.}\)
. . \(\displaystyle \text{Hence: }\,\alpha\text{ is in Quadrant 2.}\)
\(\displaystyle \tan\alpha \:=\:\dfrac{+1}{-2} \:=\:\dfrac{opp}{adj} \)
\(\displaystyle \alpha\text{ is in Quadrant 2 with: }\,opp = 1,\:adj = -2\)
. . \(\displaystyle \text{Hence: }\,hyp = \sqrt{5}\)
\(\displaystyle \sin\alpha \,=\,\dfrac{1}{\sqrt{5}},\;\cos\alpha \,=\,-\dfrac{2}{\sqrt{5}} \;\;[1]\)
\(\displaystyle \text{Since }\,\frac{3\pi}{2} < \beta < 2\pi,\;\beta\text{ is in Quadrant 4.}\)
\(\displaystyle \csc\beta \,=\,\dfrac{+6}{-5} \,=\,\dfrac{hyp}{opp}\)
\(\displaystyle \beta\text{ is in Quadrant 4 with: }\,opp = -5,\;hyp = 6\)
. . \(\displaystyle \text{Hence: }\,adj = \sqrt{11}\)
\(\displaystyle \sin\beta \,=\,-\dfrac{5}{6},\;\cos\beta \,=\,\dfrac{\sqrt{11}}{6}\;\;[2] \)
\(\displaystyle \cos(2\alpha + \beta) \;=\;\cos(2\alpha)\cos(\beta) - \sin(2\alpha)\sin(\beta) \)
. . . . . . . . . .\(\displaystyle =\;(\cos^2\!\alpha - \sin^2\!\alpha)\cos\beta - (2\sin\alpha\cos\alpha)\sin\beta\)
. . . . . . . . . .\(\displaystyle =\;\bigg[\left(\dfrac{-2}{\sqrt{5}}\right)^2 - \left(\dfrac{1}{\sqrt{5}}\right)^2\bigg]\dfrac{\sqrt{11}}{6} - \bigg[2\left(\dfrac{1}{\sqrt{5}}\right)\left(\dfrac{-2}{\sqrt{5}}\right)\bigg]\left(-\dfrac{5}{6}\right) \)
. . . . . . . . . .\(\displaystyle =\;\left(\dfrac{4}{5} - \dfrac{1}{5}\right)\left(\dfrac{\sqrt{11}}{6}\right) - \left(-\dfrac{4}{5}\right)\left(-\dfrac{5}{6}\right) \)
. . . . . . . . . .\(\displaystyle =\;\dfrac{\sqrt{11}}{10} - \dfrac{2}{3}\)
. . . . . . . . . .\(\displaystyle =\;\dfrac{3\sqrt{11} - 20}{30}\)