Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 4.4:4

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.4:4 moved to Solution 4.4:4: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The idea is first to find the general solution to the equation and then to see which angles lie between
-
<center> [[Image:4_4_4-1(3).gif]] </center>
+
<math>0^{\circ }</math>
-
{{NAVCONTENT_STOP}}
+
and
-
{{NAVCONTENT_START}}
+
<math>360^{\circ }</math>.
-
<center> [[Image:4_4_4-2(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
If we start by considering the expression
-
{{NAVCONTENT_START}}
+
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-
<center> [[Image:4_4_4-3(3).gif]] </center>
+
as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
-
{{NAVCONTENT_STOP}}
+
 
 +
 
 +
<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>
 +
 
 +
 
 +
There is then a further solution which satisfies
 +
<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
lies in the third quadrant and makes the same angle with the negative y-axis as
 +
<math>\text{1}00^{\circ }</math>
 +
makes with the positive
 +
<math>y</math>
 +
-axis, i.e.
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
makes an angle
 +
<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
 +
with the negative
 +
<math>y</math>
 +
-axis and consequently
 +
 
 +
 
 +
<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
 +
 
[[Image:4_4_4.gif|center]]
[[Image:4_4_4.gif|center]]
 +
 +
 +
There is then a further solution which satisfies
 +
<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
lies in the third quadrant and makes the same angle with the negative y-axis as
 +
<math>\text{1}00^{\circ }</math>
 +
makes with the positive
 +
<math>y</math>
 +
-axis, i.e.
 +
<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
 +
makes an angle
 +
<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
 +
with the negative
 +
<math>y</math>
 +
-axis and consequently
 +
 +
 +
<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
 +
 +
 +
FIGURE1 FIGURE2
 +
 +
Now it is easy to write down the general solution,
 +
 +
 +
<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>
 +
and
 +
 +
<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>
 +
 +
 +
and if we make
 +
<math>v</math>
 +
the subject, we get
 +
 +
 +
<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>
 +
and
 +
 +
<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>
 +
EQ6
 +
 +
For different values of the integers
 +
<math>n</math>, we see that the corresponding solutions are:
 +
 +
 +
<math>\begin{array}{*{35}l}
 +
\cdots \cdots & \cdots \cdots & \cdots \cdots \\
 +
n=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\
 +
n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\
 +
n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\
 +
n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\
 +
n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\
 +
n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\
 +
\cdots \cdots & \cdots \cdots & \cdots \cdots \\
 +
\end{array}</math>
 +
 +
 +
 +
From the table, we see that the solutions that are between
 +
<math>0^{\circ }</math>
 +
and
 +
<math>360^{\circ }</math>
 +
are
 +
 +
 +
<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }</math>
 +
and
 +
<math>v=300^{\circ }</math>

Revision as of 10:20, 1 October 2008

The idea is first to find the general solution to the equation and then to see which angles lie between 0 and 360.

If we start by considering the expression 2v+10 as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is


2v+10=110


There is then a further solution which satisfies 02v+10360, where 2v+10 lies in the third quadrant and makes the same angle with the negative y-axis as 100 makes with the positive y -axis, i.e. 2v+10 makes an angle 110−90=20  with the negative y -axis and consequently


2v+10=270−20=250



There is then a further solution which satisfies 02v+10360, where 2v+10 lies in the third quadrant and makes the same angle with the negative y-axis as 100 makes with the positive y -axis, i.e. 2v+10 makes an angle 110−90=20  with the negative y -axis and consequently


2v+10=270−20=250


FIGURE1 FIGURE2

Now it is easy to write down the general solution,


2v+10=110+n360 and

2v+10=250+n360


and if we make v the subject, we get


v=50+n180 and

v=120+n180 EQ6

For different values of the integers n, we see that the corresponding solutions are:


n=−2n=−1n=0n=1n=2n=3v=50−2180=−310v=50−1180=−130v=50+0180=50v=50+1180=230v=50+2180=410v=50+3180=590v=120−2180=−240v=120−1180=−60v=120+0180=120v=120+1180=300v=120+2180=480v=120+3180=660


From the table, we see that the solutions that are between 0 and 360 are


v=50v=120v=230 and v=300

Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.4:4"