When given a trigonometric ratio, and asked to find another ratio, we draw a right triangle as shown in the picture.
From the pythagorean theorem we find the length of AB:
[tex]|AB|= \sqrt{ AC^{2}- BC^{2} }= \sqrt{ 11^{2}- 7^{2} }= \sqrt{121-49} = \sqrt{72} = \sqrt{2*36}[/tex] [tex]= 6\sqrt{2} [/tex]
the numerical value of
cos∅=adjacent side/hypotenuse=[tex] \frac{ 6 \sqrt{2} }{11}[/tex]
since sec∅=1/cos∅, and sec∅ is negative, then cos∅ is also negative.
Answer:
cos∅=[tex]-\frac{ 6 \sqrt{2} }{11}[/tex]
