Question 6.6.5.5: finding exact Values If it is known that sinα=4/5, π/2< α......
Finding exact Values
If it is known that \sin\alpha={\frac{4}{5}},\,{\frac{\pi}{2}}\lt \alpha\lt \pi, and that \sin\beta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5},\pi\lt \beta\lt {\frac{3\pi}{2}}, find the exact value of
(a) Because \sin\alpha={\frac{4}{5}}={\frac{y}{r}} and {\frac{\pi}{2}}\lt \alpha\lt \pi, Let y = 4 and r = 5 and replace \alpha in quadrant II. The point P=(x,y)\,=\,(x,4)\,,x\,\lt 0, is on a circle of radius 5_that is, x^{2}+y^{2}=25. See Figure 27. Then
x^{2}+y^{2}=25.x^{2}+16=25. y = 4
x^{2}=25 – 16 = 9x = -3 x\lt 0
Then
\cos\alpha={\frac{x}{r}}=-{\frac{3}{5}}Alternatively, \cos \alpha can be found using identities, as follows:
\cos\alpha \underset{\overset{\uparrow }{\alpha\, \text{in quadrant II.} \cos \alpha \lt0}}{=}-{\sqrt{1-\sin^{2}\alpha}}=-{\sqrt{1-{\frac{16}{25}}}}=-{\sqrt{{\frac{9}{25}}}}=-{\frac{3}{5}}(b) Because \sin\beta={\frac{-2}{\sqrt{5}}}={\frac{y}{r}} and \pi\lt \beta\lt {\frac{3\pi}{2}}, let y = -2 and r = \sqrt{5} and replace \beta in quadrant III. The point P=\left(x,y\right)=\left(x,-2\right),x\lt 0, is on a circle of radius \sqrt{5}_that is, x^{2}+y^{2}=5. See Figure 28. Then
x^{2}+y^{2}=5x^{2}+4=5 y = -2
x^{2}=1x = -1 x\lt 0
Then
\cos\beta={\frac{x}{r}}={\frac{-1}{\sqrt{5}}}=-{\frac{\sqrt{5}}{5}}Alternatively, \cos \beta can be found using identities, as follows:
\cos\beta=-\sqrt{1-\sin^{2}\beta}=-\sqrt{1-{\frac{4}{5}}}=-\sqrt{{\frac{1}{5}}}=-{\frac{\sqrt{5}}{5}}(c) Use the results found in parts (a) and (b) and formula (1) to obtain
\begin{array}{r l}{\cos\left(\alpha+\beta\right)\,=\,\cos\alpha\cos\beta-\,\sin\alpha\sin\beta}\\ {\,}{{}=-{\frac{3}{5}}\!\left(-{\frac{\sqrt{5}}{5}}\right)-{\frac{4}{5}}\!\left(-{\frac{2{\sqrt{5}}}{5}}\right)={\frac{11{\sqrt{5}}}{25}}}\end{array}(d) {\sin\left(\alpha+\beta\right)=\sin\alpha\cos\beta+\cos\alpha\sin\beta}\\ ={\frac{4}{5}}\left(-{\frac{\sqrt{5}}{5}}\right)+\left(-{\frac{3}{5}}\right)\left(-{\frac{2{\sqrt{5}}}{5}}\right)={\frac{2{\sqrt{5}}}{25}}
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