Question 170144
<font size = 7 color = "red"><b>Stanbon's solution is correct but he does not do problem 2 and he does not explain why we know to take the positive square root in 1.

Edwin's solution:</b></font>

<pre><font size = 4 color = "indigo"><b> 
1) If {{{x}}} is an angle in quadrant 4 and {{{Cot(x)= -7/24}}}, find the value of {{{Sin}}}{{{1/2}}}{{{x}}}.

Since this involves drawing a graph in which {{{x}}} represents
the horizontal axis, not an angle,  I will temporarily change
{{{x}}} to {{{alpha)}}} to avoid a conflict of letters.  Change
the problem to read this way:

1) If {{{alpha}}} is an angle in quadrant 4 and {{{Cot(alpha)= -7/24}}}, find the value of {{{Sin}}}{{{1/2}}}{{{alpha}}}.

We must use the identity:

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((1-Cos(alpha))/2)}}}

However we do not know {{{Cos(alpha)}}}

So we must first draw the picture of the angle {{{alpha}}}:

We know that {{{Cot(alpha)}}} is by definition {{{x/y}}}, we
can draw the angle in the 4th quadrant with referent angle
is inside a triangle whose horizontal side {{{x}}} is taken
to be the numerator of {{{-7/24}}}, considered positive because
it goes right of the y-axis, and whose vertical side {{{y}}} is taken as 
the denominator {{{-24}}}, taken negative because it goes down
below the x-axis: 

{{{drawing(400,400,-30,30,-30,30, graph(400,400,-30,30,-30,30),

line(0,0,7,-24),line(7,0,7,-24),line(0,0,7,0), locate(8,-10,"y=-24"),
locate(2,4,"x=7") )}}} 

Next we calculate {{{r}}} by the Pythagorean theorem:

{{{r^2=x^2+y^2}}}
{{{r^2=(7)^2+(-24)^2}}}
{{{r^2=49+576}}}
{{{r^2=625}}}
{{{r=sqrt(625)}}}
{{{r=25}}}

So we label the slanted line segment {{{r=25}}}. 

{{{drawing(400,400,-30,30,-30,30, graph(400,400,-30,30,-30,30),
locate(.5,-18,"r=25"),
line(0,0,7,-24),line(7,0,7,-24),line(0,0,7,0), locate(8,-10,"y=-24"),
locate(2,4,"x=7") )}}} 

Now we can find {{{Cos(alpha)}}}

{{{Cos(alpha)=x/r=7/25}}}

So we substitute {{{7/25}}} for {{{Cos(alpha)}}} in

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((1-Cos(alpha))/2)}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((1-7/25)/2)}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((25/25-7/25)/2)}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((18/25)/2)}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt((18/25)(1/2))}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}{{{""+-sqrt(9/25)}}}

{{{Sin}}}{{{1/2}}}{{{alpha}}}{{{"="}}}±{{{3/5}}}

Next we must decide whether this is positive or negative:

Since {{{alpha}}} is is the 4th quadrant, then 

{{{270<alpha<360}}}° so multiplying that through by {{{1/2}}}

{{{135<1/2}}}{{{alpha<180}}}° 

The means {{{1/2}}}{{{alpha}}} is in quadrant 2.  Since
the sine is positive in the 2nd quadrant, the final answer
is 

{{{Sin}}}{{{1/2}}}{{{alpha=3/5}}} 

And of course now that we have the answer we can change
{{{alpha}}} back to {{{x}}}:

{{{Sin}}}{{{1/2}}}{{{x=3/5}}}


2) What is the solution set of the equation

 {{{Sin(2x) - Cos^2x+1 }}} = {{{ Sin^2x+ Sin(x) }}} in the interval {{{0<=x<2pi}}}?

Use the identity {{{Cos^2x= 1-Sin^2x}}} to replace {{{Cos^2x}}} on the
left side:

 {{{Sin(2x) - (1-Sin^2x) +1 }}} = {{{ Sin^2x+ Sin(x) }}}

 {{{Sin(2x) - 1+Sin^2x +1 }}} = {{{ Sin^2x+ Sin(x) }}}

{{{Sin(2x) +Sin^2x }}} = {{{ Sin^2x+ Sin(x) }}}

{{{Sin(2x)  }}} = {{{ Sin(x) }}}

{{{Sin(2x) - Sin(x)}}} = {{{0}}}

Now use identity {{{Sin(2x) = 2*Sin(x)Cos(x)}}} to
replace {{{Sin(2x)}}}

{{{2*Sin(x)Cos(x) - Sin(x)}}} = {{{0}}}

Factor out {{{Sin(x)}}}

{{{Sin(x)(2Cos(x) - 1)}}} = {{{0}}}

Use the zero-factor principle:

{{{matrix(5,4,Sin(x)=0,   "", "",   2Cos(x)-1=0,
           x = 0,      "", "",    2Cos(x)=1,     
           x = pi,     "", "",     Cos(x)=1/2,
             "",       "", "",      x=pi/3,
             "",       "", "",      x=5pi/3 )}}}

3) Find the solution set of {{{6Sin(x) + 11= 2Csc(x)}}} over the domain {{{0<x<360}}}°.

{{{6Sin(x) + 11= 2Csc(x)}}}

Use identity {{{Csc(x)=1/Sin(x)}}}


{{{6Sin(x) + 11= 2(1/Sin(x))}}}

Multiply through by {{{Sin(x)}}}

{{{6Sin^2x+11Sin(x)=2}}}

{{{6Sin^x+11Sin(x)-2=0}}}

{{{(6Sin(x)-1)(Sin(x)+2)=0}}}

Use the zero-factor principle:
 
{{{matrix(6,4,
    6Sin(x)-1=0,  "",  "",  Sin(x)+2=0,
     6Sin(x)=1,   "",  "",   Sin(x)=-2,
      Sin(x)=1/6, "",  "",   ignore,  
      "x=9.594°", "",   "",   since,
      "x=180°-9.594°", "", "", abs(Sin(x))<1,   
      "x=170.406°", "", "", "" ) }}}

Edwin</pre>