Question 1065976
</BLOCKQUOTE><FONT COLOR=red>Given that {{{theta}}} is an acute angle and {{{sin(theta) = 17/37}}}, find the exact value of sec(theta) </FONT><BR>
THE SEVENTH GRADER'S POINT OF VIEW:
If {{{theta}}} is an acute angle, it could be an angle in a right triangle, and
{{{sin(theta)=17/37}}} can be seen as a trigonometric ratio,
relating the lengths of the hypotenuse and opposite side in a right triangle:
{{{drawing(360,200,-1,35,-2,18,
green(rectangle(32.86,0,31.86,1)),
triangle(0,0,32.86,0,32.86,17),
red(arc(0,0,10,10,-27.35,0)),
locate(16,0,x),locate(33,9,17),
locate(16.43,8.5,37),locate(3,1.5,red(theta))
)}}} . With that point of view you do not need to know about quadrants,
or about trigonometric functions.
Then you would find the length of the missing side using the Pythagorean theorem:
{{{x^2+17^2=37^2}}} --> {{{x^2+289=1369}}} --> {{{x^2=1369-289}}} --> {{{x^2=1080}}} --> {{{x=sqrt(1080)=sqrt(36*30)=sqrt(36)*sqrt(30)=6sqrt(30)}}} .
Then you would find {{{cos(theta)}}} as the ratio (adjacent side)/hypotenuse:
{{{cos(theta)=highlight(6sqrt(30)/37=about0.8882)}}} (rounded).
 
THE HIGH SCHOOLER'S POINT OF VIEW:
When we need to extend the definitions of trigonometric ratios to larger angles, measuring {{{90^o}}} or more, which cannot be part of a right triangle,
we start talking about quadrants and trigonometric functions.
At that point we say that
if {{{theta}}} is an acute angle, it si in the first quadrant,
where {{{0<theta<90^o}}} or {{{0<theta<pi/2}}} ,
and all trigonometric functions are positive.
Then, we use {{{sin(theta) = 17/37}}} and {{{sin^2(theta)+cos^2(theta)=1}}}
to find {{{cos(theta)}}} :
{{{(17/37)^2+cos^2(theta)=1}}}
{{{17^2/37^2+cos^2(theta)=1}}}
{{{cos^2(theta)=1-17^2/37^2}}}
{{{cos^2(theta)=(37^2-17^2)/37^2}}}
{{{cos^2(theta)=(1369-289)/37^2}}}
{{{cos^2(theta)=1080/37^2}}}
{{{cos(theta)=sqrt(1080/37^2)=sqrt(1080)/37=highlight(6sqrt(30)/37=about0.8882)}}} (rounded).
 
THE SECOND QUESTION:
That question is beyond the 7th grade math.
At this point in his/her studies,
a student should be resigned to the fact that radians are here to stay,
because degrees are for people whose career options will be limited.
If you are there, you know that a right angle measures {{{pi/2}},
and you know where each quadrant begins and ends.
There are some trigonometric definitions and identities that
everyone dealing with trigonometry should remember:
{{{tan(theta)=sin(theta)/cos(theta)}}} ,
{{{cot(theta)=1/tan(theta)}}} ,
{{{sec(theta)=1/cos(theta)}}} ,
{{{csc(theta)=1/sin(theta)}}} , and
{{{sin^2(theta)+cos^2(theta)=1}}} .
For the rest, you should be allowed to look up the formulas.
If your life is such that you need to look up those formulas often,
you will eventually remember them.
I just search online for "trig identities" and trust Wikipedia.
It says that
{{{sin(alpha + beta)=sin(alpha)*cos(beta) + sin(beta)*cos(alpha)}}} ,
{{{cos(alpha + beta)=cos(alpha)*cos(beta) - sin(alpha)*sin(beta)}}}
{{{sin^2(theta/2)=(1-cos(theta))/2}}} , and {{{cos^2(theta)/2=(1+cos(theta))/2}}} . 
That is all I need.
I remember the exact values of the trigonometric functions for
{{{pi/4}}} and {{{pi/3}}} , because those are the angles in
the half of a square cut along the diagonal,
and an equilateral triangle:
{{{drawing(300,300,-0.1,1.1,-0.1,1.1,
rectangle(0,0,1,1),triangle(0,0,1,0,1,1),
rectangle(1,0,0.95,0.05),red(arc(0,0,0.4,0.4,-45,0)),
red(arc(0,0,0.35,0.35,-90,-45)),red(arc(1,1,0.35,0.35,90,135)),
red(arc(1,1,0.38,0.38,135,180)),locate(0.2,0.15,red(pi/4)),
locate(0.45,0,1),locate(1.03,0.55,1),
locate(0.5,0.5,sqrt(2)),rectangle(0,1,0.05,0.95)
)}}} ---> {{{sin(pi/4)=cos(pi/4)=1/sqrt(2)=sqrt(2)/2}}}
{{{system(sin(pi/3)=sqrt(1^2-(1/2)^2)=sqrt(1-1/4)=sqrt(3/4)=sqrt(3)/2,cos(pi/3)=1/2)}}} <--- {{{drawing(300,300,-0.6,0.6,-0.2,1,
triangle(-0.5,0,0.5,0,0,0.866),triangle(0,0,0.5,0,0,0.866),
rectangle(0,0,-0.05,0.05),red(arc(-0.5,0,0.3,0.3,-60,0)),
red(arc(0.5,0,0.3,0.3,-180,-120)),red(arc(0,0.866,0.3,0.3,60,120)),
locate(-0.27,0,1/2),locate(0.23,0,1/2),locate(-0.25,0.433,1),
locate(0.23,0.433,1), locate(0.03,0.4,sqrt(3)/2),
locate(-0.35,0.15,red(pi/3))
)}}}
Now, {{{2pi/3=pi-pi/3)}}} : {{{drawing(450,255,-1.2,1.2,-0.16,1.2,
grid(0),red(circle(0,0,1)),triangle(0,0,0.5,0,0.5,0.866),
triangle(0,0,-0.5,0,-0.5,0.866),locate(0.52,0.5,sin(pi/3)),
locate(-0.48,0.5,sin(2pi/3)),red(arc(0,0,0.3,0.3,-60,0)),
red(arc(0,0,0.3,0.3,-180,-120)),locate(0.15,0.2,red(pi/3)),
green(arc(0,0,0.6,0.6,-120,0)),locate(0.02,0.47,green(2pi/3))
)}}} , so {{{system(sin(2pi/3)=sin(pi/3)=sqrt(3)/2,cos(2pi/3)=-cos(pi/3)=-1/2)}}} .
Now, we can </BLOCKQUOTE><FONT COLOR=red>use the exact values for the sine and cosine of both {{{pi/4}}} and {{{2pi/3}}},
and the angle sum identity for cosine, {{{"(to"}}} {{{"find )"}}} the exact value of {{{cos(11pi/12)}}}</FONT><BR> {{{11pi/12=(8+3)pi/12=8pi/12+3pi/12=2pi/3+pi/4}}} ,
so applying the sum identity for cosine
{{{cos(11pi/12)=cos(2pi/3)*cos(pi/4)-sin(2pi/3)*sin(pi/4)=(-1/2)(sqrt(2)/2)-(sqrt(3)/2)(sqrt(2)/2)=(-sqrt(2)-sqrt(2)*sqrt(3))/4=highlight(-sqrt(2)(1+sqrt(3))/4=-(sqrt(2)+sqrt(6))/4)}}}
 
As for using </BLOCKQUOTE><FONT COLOR=red>the exact value of {{{cos(11pi/6)}}} and the half-angle identity for sine to find the exact value of {{{sin(11pi/12)}}}</FONT><BR> ,
we would have to figure out how to find {{{cos(11pi/6)}}} first.
And, why would we do that, instead of using the already found value of {{{cos(11pi/12)}}} ,
and the trigonometric identity {{{sin^2(theta)+cos^2(theta)=1}}} ?
Well, It is not that hard.
{{{11pi/6=(12-1)pi/6=12pi/6-pi/6=2pi-pi/6}}}, so {{{drawing(255,450,-0.2,1.16,-1.2,1.2,
grid(0),red(circle(0,0,1)),triangle(0,0,0.866,0.5,0.866,-0.5),
rectangle(0.866,0,0.816,0.05),red(arc(0,0,0.6,0.6,-30,0)),
red(arc(0,0,0.4,0.4,0,30)),locate(0.3,0.2,red(pi/6)),
green(arc(0,0,0.35,0.35,30,360)),locate(0.05,-0.16,green(11pi/6)),
locate(0.433,0.25,1),locate(0.87,0.3,sqrt(3)/2),locate(0.45,0,1/2)
)}}} , and we see that
{{{system(sin(11pi/6)=-sin(pi/6)=-1/2,cos(11pi/6)=cos(pi/6)=sqrt(3)/2)}}}
Using that, and the half-angle identity for sine, {{{sin^2(theta/2)=(1-cos(theta))/2}}} ,
{{{sin^2(11pi/12)=sin^2((11pi/6)/2)=(1-cos(11pi/6))/2=(1-sqrt(3)/2)/2=(2-sqrt(3))/4}}} ,
and since {{{11pi/6}}} was almost a full turn, almost at the far end of the fourth quadrant,
we know that half of that, {{{11pi/12}}}, would be almost,
almost at the end of the second quadrant.
We also know that in the second quadrant sine is positive, and cosine is negative.
So, {{{sin(11pi/12)=sqrt((2-sqrt(3))/4)=highlight(sqrt(2-sqrt(3))/2))}}} .