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If sin(x) = 1/3 and sec(y) = 5/4 , where x and y lie between 0 and pie/2, evaluate the expression cos(x+y).
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ok you have\[sin(x)=\frac{1}{3}\] and you are going to need \[cos(x)=\frac{\sqrt{8}}{3}\] by pythatoras
you also have \[sec(y)=\frac{5}{4}\] which means \[cos(y)=\frac{4}{5}\]
you also need \[sin(y)=\frac{3}{5}\] also by pathagoras
now \[cos(x+y)=cos(x)cos(y)-sin(x)sin(y)\] and now all these numbers are known so you substitute then into the formula
\[\frac{\sqrt{8}}{3}\times \frac{4}{5}-\frac{1}{3}\times \frac{3}{5}\]
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=\[\frac{8\sqrt{2}-3}{15}\]
could you explain the pythagoras part for me? how is sin x = 1/3 is cos x = sqrt 8 /3. could you please show that work?
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