Question 189534
<pre><font size = 4 color = "indigo"><b>
I can do your 2nd problem but not the 1st, for you've 
made a mistake in copying the first problem. Here's why:

sin<font face = "symbol">q</font> = 12/13, 3<font face = "symbol">p</font>/2 < <font face = "symbol">q</font> < 2<font face = "symbol">p</font>. Find sin2<font face = "symbol">q</font>, cos2<font face = "symbol">q</font>, tan2<font face = "symbol">q</font>?

3<font face = "symbol">p</font>/2 < <font face = "symbol">q</font> < 2<font face = "symbol">p</font>

tells us that <font face = "symbol">q</font> is in quadrant IV.
But the sine is negative in quadrant IV, yet you have 
sin<font face = "symbol">q</font> = 12/13, which is a positive number.
That cannot be.

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Here's your 2nd problem:

cos<font face = "symbol">q</font> = -1/4,csc<font face = "symbol">q</font> > 0, find sin2<font face = "symbol">q</font>,cos2<font face = "symbol">q</font>, tan2<font face = "symbol">q</font>

This tells us that <font face = "symbol">q</font> is in quadrant II,
because that is the only quadrant in which the cosine is negative and
the cosecant is positive.

So we draw an angle in Quadrant II to represent angle <font face = "symbol">q</font> 
{{{drawing(400,400,-3,3,-1,5,
graph(400,400,-3,3,-1,5,sqrt(1-x^2)*sqrt(.25+x)/sqrt(.25+x)), line(0,0,-1,sqrt(15)), locate(1/2,1.2,"theta") )}}}

Next we draw a perpendicular to the x axis.  Since the cosine
is {{{x/r}}}, we make the x value -1, which is the numerator of 
{{{(-1)/4}}} and make the hypotenuse or the r equal to the the 
denominator of {{{(-1)/4}}} which is 4: 

{{{drawing(400,400,-3,3,-1,5,line(0,0,-1,sqrt(15)),
locate(-.8,.3,"x=-1"), locate(-1/2, 2.5,"r=4"),
graph(400,400,-3,3,-1,5,sqrt(1-x^2)*sqrt(.25+x)/sqrt(.25+x)), triangle(0,0,-1,0,-1,sqrt(15)), locate(1/2,1.2,"theta") )}}}

Next we calculate the value of y by the Pythagorean
theorem equation

{{{x^2+y^2=r^2}}}
{{{(-1)^2+y^2=(4)^2}}}
{{{1+y^2=16}}}
{{{y^2=15}}}
{{{y=sqrt(15)}}}

So we label the y-value with that

{{{drawing(400,400,-3,3,-1,5, line(0,0,-1,sqrt(15)),
locate(-.8,.3,"x=-1"), locate(-1/2, 2.5,"r=4"),
graph(400,400,-3,3,-1,5,sqrt(1-x^2)*sqrt(.25+x)/sqrt(.25+x)), triangle(0,0,-1,0,-1,sqrt(15)), locate(1/2,1.2,"theta"),locate(-1.7,1.5,y=sqrt(15)) )}}}


Now we can find sin2<font face = "symbol">q</font> using the
identity:

sin2<font face = "symbol">q</font> = 2sin<font face = "symbol">q</font>cos<font face = "symbol">q</font>

because we now know that sin<font face = "symbol">q</font> = {{{y/r}}} = {{{sqrt(15)/4}}}, and cos<font face = "symbol">q</font> = {{{x/r}}} = {{{-1/4}}}.

sin2<font face = "symbol">q</font> = 2sin<font face = "symbol">q</font>cos<font face = "symbol">q</font> = {{{2(sqrt(15)/4)(-1/4)=-2sqrt(15)/16=-sqrt(15)/8}}}

Next we find cos2<font face = "symbol">q</font> using any one of
these three identities:

cos2<font face = "symbol">q</font> = cos<sup>2</sup><font face = "symbol">q</font> - sin<sup>2</sup><font face = "symbol">q</font>
cos2<font face = "symbol">q</font> = 2cos<sup>2</sup><font face = "symbol">q</font> - 1
cos2<font face = "symbol">q</font> = 1 - 2sin<sup>2</sup><font face = "symbol">q</font>.

I'll choose the second one, although any one will give you the
same answer:

cos2<font face = "symbol">q</font> = 2cos<sup>2</sup><font face = "symbol">q</font> - 1
cos2<font face = "symbol">q</font> = 2{{{(-1/4)^2}}} - 1
cos2<font face = "symbol">q</font> = 2{{{1/16}}} - 1
cos2<font face = "symbol">q</font> = {{{2/16}}} - 1
cos2<font face = "symbol">q</font> = {{{1/8}}} - 1
cos2<font face = "symbol">q</font> = {{{-7/8}}}

Finally we can find tan2<font face = "symbol">q</font> from
the identity

tan2<font face = "symbol">q</font> = sin2<font face = "symbol">q</font>/cos2<font face = "symbol">q</font>

tan2<font face = "symbol">q</font> = {{{(-sqrt(15)/8)/(-7/8) = (-sqrt(15)/8)(-8/7) = sqrt(15)/7}}}

Edwin</pre>