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File:Y=csc(x).gif The graph of y=csc(x) in radians
File:Y=sec(x).gif The graph of y=sec(x) in radians
File:Y=cot(x).gif The graph of y=cot(x) in radians
Aside from the classic 3 trigonmetical functions, there are now 3 more you must be aware of; the reciprocals of our standard ones. We have the cosecant (csc), secant (sec), and cotangent (cot). These are defined as:
csc
θ
=
1
sin
θ
{\displaystyle \csc \theta ={\frac {1}{\sin \theta }}}
sec
θ
=
1
cos
θ
{\displaystyle \sec \theta ={\frac {1}{\cos \theta }}}
cot
θ
=
1
tan
θ
=
cos
θ
sin
θ
{\displaystyle \cot \theta ={\frac {1}{\tan \theta }}={\frac {\cos \theta }{\sin \theta }}}
Each of these is undefined for certain values of
θ
{\displaystyle \theta }
. For example; cscθ is undefined when θ=0,180,360..., because sinθ=0 at these points.
Each of the graphs of these functions all have asymptotes intervals of 180 degrees.
Using our new definitions of reciprocal functions, we are able to obtain 2 new identities based of Pythagoras' theorem.
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
Dividing both sides by
cos
2
θ
{\displaystyle \cos ^{2}\theta }
sin
2
θ
cos
2
θ
+
cos
2
θ
cos
2
θ
=
1
cos
2
θ
{\displaystyle {\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}+{\frac {\cos ^{2}\theta }{\cos ^{2}\theta }}={\frac {1}{\cos ^{2}\theta }}}
tan
2
θ
+
1
=
sec
2
θ
{\displaystyle \tan ^{2}\theta +1=\sec ^{2}\theta }
There is also a second identity:
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
Dividing both sides by
sin
2
θ
{\displaystyle \sin ^{2}\theta }
sin
2
θ
sin
2
θ
+
cos
2
θ
sin
2
θ
=
1
sin
2
θ
{\displaystyle {\frac {\sin ^{2}\theta }{\sin ^{2}\theta }}+{\frac {\cos ^{2}\theta }{\sin ^{2}\theta }}={\frac {1}{\sin ^{2}\theta }}}
1
+
cot
2
θ
=
csc
2
θ
{\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta }
Question 1:'Find cosec 120, leaving your answer in surd form'
Solution:
csc
120
=
1
s
i
n
120
{\displaystyle \csc 120={\frac {1}{sin120}}}
=
1
/
3
2
{\displaystyle =1/{\frac {\sqrt {3}}{2}}}
=
2
3
{\displaystyle ={\frac {2}{\sqrt {3}}}}
Question 2:'Find all values of x in the interval 0≤x≤360 for:
sec
2
x
=
4
+
2
tan
x
{\displaystyle \sec ^{2}x=4+2\tan x}
Solution:
sec
2
x
=
4
+
2
tan
x
{\displaystyle \sec ^{2}x=4+2\tan x}
tan
2
x
+
1
=
4
+
2
tan
x
{\displaystyle \tan ^{2}x+1=4+2\tan x}
tan
2
x
−
2
tan
x
−
3
=
0
{\displaystyle \tan ^{2}x-2\tan x-3=0}
(
tan
x
−
3
)
(
tan
x
+
1
)
=
0
{\displaystyle (\tan x-3)(\tan x+1)=0}
tan
x
=
3
o
r
tan
x
=
−
1
{\displaystyle \tan x=3or\tan x=-1}
If
tan
x
=
3
{\displaystyle \tan x=3}
x
=
71.6
,
251.6
{\displaystyle x=71.6,251.6}
If
tan
x
=
−
1
{\displaystyle \tan x=-1}
x
=
135
,
315
{\displaystyle x=135,315}
x
=
71.6
,
135
,
251.6
,
315
{\displaystyle x=71.6,135,251.6,315}