Question 289398
Find {{{ sin2x }}} if {{{ secx=-4 }}} and {{{ sinx<0 }}}

<pre><font size = 4 color = "indigo"><b>
Since the secant and the sine of x are both negative, that means
x is in the third quadrant. between 180° and 270° in degress,
or between {{{pi}}} and {{{3pi/2}}} in radians.  So therefore
2x will be between 360° and 540° which is actually the 1st and
2nd quadrants.  The sine is positive in both the 1st and 2nd
quadrants, so the answer will be posotive.

We need these identities:

1. {{{Sin(2theta) = 2Sin(theta)Cos(theta)}}}
2. {{{Cos^2theta + Sin^2theta=1}}} 
3. {{{Cos(theta)=1/Sec(theta)}}} 
 
First we'll find {{{Cos(x)}}} using 3.

{{{Cos(x)=1/Sec(x)=1/(-4)=-1/4}}}

Next we'll find {{{Sin(x)}}} using 2.

{{{Cos^2x + Sin^2x=1}}}

{{{(-1/4)^2+Sin^2x=1}}}
 
{{{1/16+Sin^2x=1}}}

{{{Sin^2x=1-1/16}}}

{{{Sin^2x=16/16-1/16}}}

{{{Sin^2x=15/16}}}

{{{Sin(x)=""+-sqrt(15/16)=""+-sqrt(15)/4}}}

However, since {{{Sin(x)<0}}} we take the negative sign,

{{{Sin(x)=-sqrt(15)/4}}}

Now we can substitute in 1:

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

{{{Sin(2x) = 2(-sqrt(15)/4)(-1/4)}}}

{{{Sin(2x) = 2sqrt(15)/16)}}}

{{{Sin(2x) = sqrt(15)/8}}}

Edwin</pre>