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# help !!!!

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If sin x = 2 3 and sec y = 5 3 , where x and y lie between 0 and π/2, evaluate sin(x + y).

Feb 1, 2020

#1
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I assume you mean:

$$sin(x)=\frac{2}{3}$$

$$sec(y)=\frac{5}{3}$$

x and y lie between $$0$$ and $$\frac{\Pi}{2}$$ Radians

Well let's convert it to degrees

so x and y lie between 0 degrees and 90 degrees.

Thus

$$sin(x)=\frac{2}{3}$$

Take arcsin

$$x=sin^{-1}(\frac{2}{3})$$

$$x=41.81$$ degrees which lie between 0 degrees and 90 degrees

$$sec(y)=\frac{5}{3}$$

$$y=sec^{-1}(\frac{5}{3})$$

$$y=53.13$$ degrees which lie between 0 degrees and 90 degrees

$$sin(x+y)=sin(x)cos(y)+sin(y)cos(x)$$

applying this formula

Feb 1, 2020
#2
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$$\displaystyle \sin x = 2/3, \text{ so }\cos^{2}x=1-(2/3)^{2}=5/9, \text{ so }\cos x =\sqrt{5}/3.$$

$$\displaystyle \sec y = 5/3, \text{ so } \cos y = 3/5, \text{ so } \sin y = 4/5.$$

$$\displaystyle \sin(x+y)=\sin x.\cos y+\cos x.\sin y = \frac{2}{3}.\frac{3}{5}+\frac{\sqrt{5}}{3}.\frac{4}{5}=\frac{6+4\sqrt{5}}{15}\approx 0.99628.$$

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Feb 2, 2020