cos x -tan x=
pls answer
Answers
Answer:
Looking at the graphs of
y
=
x
and
y
=
cos
x
we see that there is exactly one real solution, actually somewhere in
(
1
2
,
1
)
:
graph{(y-x)(y-cos x) = 0 [-5, 5, -2.5, 2.5]}
Typically for such an equation with mixed polynomial and trigonometric terms, there is no algebraic solution.
We can use Newton's method to get a sequence of increasingly better approximations.
Let:
f
(
x
)
=
x
−
cos
x
Then:
f
'
(
x
)
=
1
+
sin
x
Newton's method tells us that if we have an approximation
a
i
to a zero of
f
(
x
)
then a better approximation is given by:
a
i
+
1
=
a
i
−
f
(
a
i
)
f
'
(
a
i
)
Choosing
a
0
=
1
as our first approximation, we find:
a
1
=
a
0
−
a
0
−
cos
a
0
1
+
sin
a
0
≈
0.75036386784
a
2
=
a
1
−
a
1
−
cos
a
1
1
+
sin
a
1
≈
0.73911289091
a
3
=
a
2
−
a
2
−
cos
a
2
1
+
sin
a
2
≈
0.73908513339
a
4
=
a
3
−
a
3
−
cos
a
3
1
+
sin
a
3
≈
0.73908513322
a
5
=
a
4
−
a
4
−
cos
a
4
1
+
sin
a
4
≈
0.73908513322
Answer:
Use Newton's method to find:
x
≈
0.73908513322
Explanation:
Looking at the graphs of
y
=
x
and
y
=
cos
x
we see that there is exactly one real solution, actually somewhere in
(
1
2
,
1
)
:
graph{(y-x)(y-cos x) = 0 [-5, 5, -2.5, 2.5]}
Typically for such an equation with mixed polynomial and trigonometric terms, there is no algebraic solution.
We can use Newton's method to get a sequence of increasingly better approximations.
Let:
f
(
x
)
=
x
−
cos
x
Then:
f
'
(
x
)
=
1
+
sin
x
Newton's method tells us that if we have an approximation
a
i
to a zero of
f
(
x
)
then a better approximation is given by:
a
i
+
1
=
a
i
−
f
(
a
i
)
f
'
(
a
i
)
Choosing
a
0
=
1
as our first approximation, we find:
a
1
=
a
0
−
a
0
−
cos
a
0
1
+
sin
a
0
≈
0.75036386784
a
2
=
a
1
−
a
1
−
cos
a
1
1
+
sin
a
1
≈
0.73911289091
a
3
=
a
2
−
a
2
−
cos
a
2
1
+
sin
a
2
≈
0.73908513339
a
4
=
a
3
−
a
3
−
cos
a
3
1
+
sin
a
3
≈
0.73908513322
a
5
=
a
4
−
a
4
−
cos
a
4
1
+
sin
a
4
≈
0.73908513322
So you can see that the approximations converge quite rapidly.