if tan theta is equal to 5 by 6 and the terminal side of the quadrant is in the third quadrant find the remaining trigonometric ratio of theta.
Answers
Answer:
Here,
0
<
tan
x
=
5
12
.
So,
x
∈
Q
1
and
=
k
π
+
arctan
(
5
12
)
,
k
=
0
,
2
,
4
,
6
,
.
.
or
∈
Q
3
and
=
k
π
+
arctan
(
5
12
)
,
k
=
1
.
3
,
5
,
...
.
Easily,
sin
x
=
5
±
√
12
2
+
5
2
=
±
5
13
and
cos
x
=
±
√
1
−
sin
2
x
=
±
12
13
If
x
∈
(all positive )
Q
1
, choose positive values.
Otherwise, both are negative.
If x is chosen in
(
0
,
2
π
)
,
it is either
arctan
(
3
12
)
=
22.62
o
=
0.3948
r
a
d
, nearly, or
180
o
+
22.62
o
=
202.62
o
=
3.5364
r
a
d
, nearly.
See the combined graph of
y
=
tan
x
,
y
=
sin
x
and
y
=
cos
x
,
depicting all these aspects. Slide the graph
←
and
→
, to see
x-solutions, as the meet of
y
=
5
12
with the periodic graph
y
=
tan
x
, in infinitude. The circular dots give the answer as y-
values, respectively.
graph{(y-tan x) ( y- sin x ) ( y- cos x )(y- 5/12+0x)(x-0.395+0.001 y)(x-3.536 + 0.0001 y) ((x-0.395)^2+(y-12/13)^2-0.001)((x-3.536)^2+(y+12/13)^2 - 0.001)((x-0.395)^2+(y-5/13)^2-0.001)((x-3.536)^2+(y+5/13)^2-0.001)=0[0 5 -1.2 1-2]}