If sin x = 2 3 and sec y = 5 3 , where x and y lie between 0 and π/2, evaluate sin(x + y).
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
we get that sin(41.81+53.13)=0.99628 radians..
\(\displaystyle \sin x = 2/3, \text{ so }\cos^{2}x=1-(2/3)^{2}=5/9, \text{ so }\cos x =\sqrt{5}/3.\)
\(\displaystyle \sec y = 5/3, \text{ so } \cos y = 3/5, \text{ so } \sin y = 4/5.\)
\(\displaystyle \sin(x+y)=\sin x.\cos y+\cos x.\sin y = \frac{2}{3}.\frac{3}{5}+\frac{\sqrt{5}}{3}.\frac{4}{5}=\frac{6+4\sqrt{5}}{15}\approx 0.99628.\)
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