Question 323917
<pre><b>
If sin a = 0.2 then cos a must equal?

Since {{{sine}}}{{{""=""}}}{{{(opposite)/(hypotenuse)}}} and since

{{{0.2}}}{{{""=""}}} the fraction {{{0.2/1}}} if we draw a right triangle
with one side equal to the numerator {{{0.2}}} of that fraction and the
hypotenuse equal to the denominator of that fraction {{{1}}}, then the
angle that the side that equals {{{0.2}}} is opposite to will be angle a:

{{{drawing(400,1200/7,-1,6,-1,2, triangle(0,0,sqrt(24),0,sqrt(24),1),
locate(1,.25,a), rectangle(sqrt(24)-.15,0,sqrt(24),.15), 
locate(sqrt(24)+.1,.6,0.2), locate(3,.9,1), arc(0,0,2.8,-2.8,0,11.53695903)
)}}}
 
Since {{{cosine}}}{{{""=""}}}{{{(adjacent)/(hypotenuse)}}} all we need
do is calculate the side adjacenet to angle a, which is the bottom side,
call it x.

{{{drawing(400,1200/7,-1,6,-1,2, triangle(0,0,sqrt(24),0,sqrt(24),1),
locate(1,.25,a), rectangle(sqrt(24)-.15,0,sqrt(24),.15), 
locate(sqrt(24)+.1,.6,0.2), locate(3,.9,1), arc(0,0,2.8,-2.8,0,11.53695903),
locate(2.5,0,x)
)}}}

We use the Pythagorean theorem to calculate x:

{{{adjacent^2+opposite^2=hypotenuse^2}}}
{{{x^2+(0.2)^2=1^2}}}
{{{x^2+0.04=1}}}
{{{x^2=.96}}}
{{{x^2=96/100}}}
{{{x^2=24/25}}}
{{{x=sqrt(24/25)}}}
{{{x=sqrt(24)/sqrt(25)}}}
{{{x=sqrt(24)/5}}}
{{{x=sqrt(4*6)/5}}}
{{{x=2sqrt(6)/5}}}

So {{{cos(a)}}}{{{""=""}}}{{{adjacent/(hypotenuse)}}}{{{"="}}}{{{(2sqrt(6)/5)/1}}}{{{""=""}}}{{{2sqrt(6)/5}}}

The decimal value of that is 0.9797958971, and you can get that
just by using your TI-84 calculator

cos(sin<sup>-1</sup>(0.2))
     .9797958971  

Edwin</pre>