Question 203931


<font size = 8 color = "red"><b>Edwin's solution:</b></font>


<pre><font size = 4 color = "indigo"><b>
a) {{{Sin(x)=12/13}}} and {{{Cos(x)<0}}}

The sine is positive and the cosine is negative
only in quadrant II

To find {{{Cos(x)}}}:

{{{Sin^2x+Cos^2x=1}}} <-- fundamental identity to use
{{{Cos^2x=1-Sin^2x}}}
{{{Cos^2x=1-(12/13)^2}}}
{{{Cos^2x=1-144/169}}}
{{{Cos^2x=169/169-144/169}}}
{{{Cos^2x=25/169}}}
{{{Cos(x)=""+-sqrt(25/169)}}}

But we are given that {{{Cos(x)<0}}}
so we choose the negative answer for {{{Cos(x)}}},
therefore:
{{{Cos(x)=-sqrt(25/169)}}}
{{{Cos(x)=-5/13)}}}

------------------

To find {{{Tan(x)}}}:

{{{Tan(x)=Sin(x)/Cos(x)}}}   <-- fundamental identity to use.

{{{Tan(x)=Sin(x)/Cos(x)=(12/13)/(-5/13)=(12/13)*(-13/5)=(12/cross(13))*(-cross(13)/5)=-12/5}}}

------------------

To find {{{Sec(x)}}}

{{{Sec(x) = 1/Cos(x)}}}   <--- fundamental identity to use

{{{Sec(x) = 1/(-5/13) = 1*(-13/5) = -13/5}}}

-------------------

To find {{{Csc(x)}}}

{{{Csc(x) = 1/Sin(x)}}}   <--- fundamental identity to use

{{{Csc(x) = 1/(12/13) = 1*(13/12) = 13/12}}}

==================================================

b){{{Cos(x) =-4/3}}} , {{{Sin(x)>0}}}

Sorry this is a mistake because {{{-4/3=-1&1/3}}} and 
cosines are always between {{{-1}}} and {{{1}}}.

So if you didn't copy it wrong, then there
is no solution!

=================================================

c) {{{Tan(x) = -24/7}}} , {{{Sin(x)>0}}}

The tangent is negative and the sine is positive
only in quadrant II

To find {{{Sec(x)}}}:

{{{1+Tan^2x=Sec^2x}}} <-- fundamental identity to use
{{{Sec^2x=1+Tan^2x}}}
{{{Sec^2x=1+(-24/7)^2}}}
{{{Sec^2x=1+576/49}}}
{{{Sec^2x=49/49+576/49}}}
{{{Sec^2x=625/49}}}
{{{Sec(x)=""+-sqrt(625/49)}}}

But since this is in Quadrant II, the secant, which
is the reciprocal of the cosine, is negative, we choose 
the negative answer for {{{Sec(x)}}}, therefore:

{{{Sec(x)=-sqrt(625/49)}}}
{{{Sec(x)=-25/7)}}}

------------------

To find {{{Cos(x)}}}:

{{{Sec(x)=1/Cos(x)}}}   <-- fundamental identity to use.

{{{Sec(x)Cos(x)=1}}}

{{{Cos(x)=1/Sec(x)=1/(-25/7)=1*(-7/25)=-7/25}}}

------------------
To find {{{Sin(x)}}}:

{{{Sin^2x+Cos^2x=1}}} <-- fundamental identity to use
{{{Sin^2x=1-Cos^2x}}}
{{{Sin^2x=1-(-7/25)^2}}}
{{{Sin^2x=1-49/625}}}
{{{Sin^2x=625/625-49/625}}}
{{{Sin^2x=576/625}}}
{{{Sin(x)=""+-sqrt(576/625)}}}

But we are given that {{{Sin(x)>0}}}
so we choose the positive answer for {{{Sin(x)}}},
therefore:
{{{Sin(x)=sqrt(576/625)}}}
{{{Sin(x)=24/25)}}}

-------------------

To find {{{Csc(x)}}}

{{{Csc(x) = 1/Sin(x)}}}   <--- fundamental identity to use

{{{Csc(x) = 1/(24/25) = 1*(25/24) = 25/24}}}

-------------------

To find {{{Cot(x)}}}:

{{{Cot(x)=1/Tan(x)}}}   <-- fundamental identity to use.

{{{Cot(x)=1/Tan(x)=1/(-24/7)=1*(-7/24)=-7/24}}}

==================================================

d) {{{Tan(x) = 4/3}}}, {{{Sin(x)>0}}}

The tangent is positive and the sine is positive
only in quadrant I, so all trigonometric ratios are
positive:

To find {{{Sec(x)}}}:

{{{1+Tan^2x=Sec^2x}}} <-- fundamental identity to use
{{{Sec^2x=1+Tan^2x}}}
{{{Sec^2x=1+(4/3)^2}}}
{{{Sec^2x=1+16/9}}}
{{{Sec^2x=9/9+16/9}}}
{{{Sec^2x=25/9}}}
{{{Sec(x)=""+-sqrt(25/9)}}}

But since this is in Quadrant I, all trig rations are
positive therefore:

{{{Sec(x)=sqrt(25/9)}}}
{{{Sec(x)=5/3)}}}

------------------

To find {{{Cos(x)}}}:

{{{Sec(x)=1/Cos(x)}}}   <-- fundamental identity to use.

{{{Sec(x)Cos(x)=1}}}

{{{Cos(x)=1/Sec(x)=1/(5/3)=1*(3/5)=3/5}}}

------------------
To find {{{Sin(x)}}}:

{{{Sin^2x+Cos^2x=1}}} <-- fundamental identity to use
{{{Sin^2x=1-Cos^2x}}}
{{{Sin^2x=1-(3/5)^2}}}
{{{Sin^2x=1-9/25}}}
{{{Sin^2x=25/25-9/25}}}
{{{Sin^2x=16/25}}}
{{{Sin(x)=""+-sqrt(16/25)}}}

But we are given that {{{Sin(x)>0}}}
so we choose the positive answer for {{{Sin(x)}}},
therefore:
{{{Sin(x)=sqrt(16/25)}}}
{{{Sin(x)=4/5)}}}

-------------------

To find {{{Csc(x)}}}

{{{Csc(x) = 1/Sin(x)}}}   <--- fundamental identity to use

{{{Csc(x) = 1/(4/5) = 1*(5/4) = 5/4}}}

-------------------

To find {{{Cot(x)}}}:

{{{Cot(x)=1/Tan(x)}}}   <-- fundamental identity to use.

{{{Cot(x)=1/Tan(x)=1/(3/4)=1*(4/3)=4/3}}}

-------------------------------

e) {{{Cot(x)= 4}}} , sinx>0

The cotangent is positive and the sine is positive
only in quadrant I, so all trigonometric ratios are
positive:

To find {{{Csc(x)}}}:

{{{1+Cot^2x=Csc^2x}}} <-- fundamental identity to use
{{{Csc^2x=1+Cot^2x}}}
{{{Sec^2x=1+(4)^2}}}
{{{Sec^2x=1+16}}}
{{{Sec^2x=17}}}
{{{Sec(x)=""+-sqrt(17)}}}

But since this is in Quadrant I, all trig ratios are
positive therefore:

{{{Sec(x)=sqrt(17)}}}

------------------

To find {{{Cos(x)}}}:

{{{Sec(x)=1/Cos(x)}}}   <-- fundamental identity to use.

{{{Sec(x)Cos(x)=1}}}

{{{Cos(x)=1/Sec(x)=1/sqrt(17)=(1/sqrt(17))*(sqrt(17)/sqrt(17))= sqrt(17)/17 }}}

------------------
To find {{{Sin(x)}}}:

{{{Sin^2x+Cos^2x=1}}} <-- fundamental identity to use
{{{Sin^2x=1-Cos^2x}}}
{{{Sin^2x=1-(sqrt(17)/17)^2}}}
{{{Sin^2x=1-17/289}}}
{{{Sin^2x=289/289-17/289}}}
{{{Sin^2x=272/289}}}
{{{Sin^2x=16/17}}}
{{{Sin(x)=""+-sqrt(16/17)}}}

But we are given that {{{Sin(x)>0}}}
so we choose the positive answer for {{{Sin(x)}}},
therefore:
{{{Sin(x)=sqrt(16/17)}}}
{{{Sin(x)=4/sqrt(17)=(4/sqrt(17))(sqrt(17)/sqrt(17))=(4sqrt(17))/17}}}

-------------------

To find {{{Csc(x)}}}

{{{Csc(x) = 1/Sin(x)}}}   <--- fundamental identity to use

{{{Csc(x) = 1/(4/sqrt(17)) = 1*(sqrt(17)/4) = sqrt(17)/4}}}

-------------------

To find {{{Cot(x)}}}:

{{{Cot(x)=1/Tan(x)}}}   <-- fundamental identity to use.

{{{Cot(x)=1/Tan(x)=1/((4))=1/4}}}

=====================================================

f) {{{Cos(x)=3/5}}} 

You didn't give enough information, as you need to be
given that another trig ratio is >0 or <0.

Edwin</pre>