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θ を解く
\theta =2\pi n_{2}+\frac{\pi }{3}\text{, }n_{2}\in \mathrm{Z}<br/>\theta =2\pi n_{3}+\frac{5\pi }{3}\text{, }n_{3}\in \mathrm{Z}<br/>\theta =\pi n_{1}\text{, }n_{1}\in \mathrm{Z}
θ
=
2
π
n
2
+
3
π
,
n
2
∈
Z
θ
=
2
π
n
3
+
3
5
π
,
n
3
∈
Z
θ
=
π
n
1
,
n
1
∈
Z
グラフ
クイズ
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
θ
cos
θ
=
2
sin
θ
Web 検索からの類似の問題
Why do we lose solutions when we divide 2\sin\theta\cos\theta=\sin\theta by \sin\theta?
Why do we lose solutions when we divide
2
sin
θ
cos
θ
=
sin
θ
by
sin
θ
?
https://math.stackexchange.com/q/2806858
Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by 0 is not valid. So going from 2 \sin \theta \cos \theta = \sin \theta ...
Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by
0
is not valid. So going from
2
sin
θ
cos
θ
=
sin
θ
...
Simple Trig Equations - Why is it Wrong to Cancel Trig Terms?
Simple Trig Equations - Why is it Wrong to Cancel Trig Terms?
https://math.stackexchange.com/questions/1416619/simple-trig-equations-why-is-it-wrong-to-cancel-trig-terms
The two issues are: When cancelling a factor, note that this is only possible when the factor is not zero; but the factor may be zero in the solution to the problem. In this case \sin\theta = 0 ...
The two issues are: When cancelling a factor, note that this is only possible when the factor is not zero; but the factor may be zero in the solution to the problem. In this case
sin
θ
=
0
...
How do I simplify this expression further?
How do I simplify this expression further?
https://math.stackexchange.com/questions/472115/how-do-i-simplify-this-expression-further
You have \frac{1}{x^2+y^2} = 1+\left|\frac{2}{y^2/x^2+1}-1\right| = 1+\frac{|x^2-y^2|}{x^2+y^2}, or put differently 1 = x^2+y^2 + |x^2-y^2|. At this point solving by cases helps. Where |x|<|y| ...
You have
x
2
+
y
2
1
=
1
+
∣
∣
∣
∣
y
2
/
x
2
+
1
2
−
1
∣
∣
∣
∣
=
1
+
x
2
+
y
2
∣
x
2
−
y
2
∣
,
or put differently
1
=
x
2
+
y
2
+
∣
x
2
−
y
2
∣
.
At this point solving by cases helps. Where
∣
x
∣
<
∣
y
∣
...
Using the exponential form of a complex number and De Moivre's theorem
Using the exponential form of a complex number and De Moivre's theorem
https://math.stackexchange.com/questions/1523448/using-the-exponential-form-of-a-complex-number-and-de-moivres-theorem
As you noted, \cos(2\theta+\pi/2) = \operatorname{Re}(e^{i(2\theta+\pi/2)}). e^{i(2\theta+\pi/2)}=(e^{i\theta})^2e^{i\pi/2}=(\cos\theta+i\sin\theta)^2(i) =i(\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta) ...
As you noted,
cos
(
2
θ
+
π
/
2
)
=
R
e
(
e
i
(
2
θ
+
π
/
2
)
)
.
e
i
(
2
θ
+
π
/
2
)
=
(
e
i
θ
)
2
e
i
π
/
2
=
(
cos
θ
+
i
sin
θ
)
2
(
i
)
=
i
(
cos
2
θ
+
2
i
cos
θ
sin
θ
−
sin
2
θ
)
...
Is the point of a shape with the greatest average ray length also the “centroid”?
Is the point of a shape with the greatest average ray length also the “centroid”?
https://math.stackexchange.com/questions/1588150/is-the-point-of-a-shape-with-the-greatest-average-ray-length-also-the-centroid
I have confused you with the definition of mean radius with the average ray length. When people think of average radius they think of "average squared minimized" or mean radius. What I am focusing on ...
I have confused you with the definition of mean radius with the average ray length. When people think of average radius they think of "average squared minimized" or mean radius. What I am focusing on ...
Do we have to show it for both cases?
Do we have to show it for both cases?
https://math.stackexchange.com/questions/1459841/do-we-have-to-show-it-for-both-cases
You can choose C=c+\pi, and then \sin{(\theta+C)}=-\sin{(\theta+c)}, so having the \pm there doesn't create any more solutions if you allow any c \in [0,2\pi).
You can choose
C
=
c
+
π
, and then
sin
(
θ
+
C
)
=
−
sin
(
θ
+
c
)
, so having the
±
there doesn't create any more solutions if you allow any
c
∈
[
0
,
2
π
)
.
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例
二次方程式の公式
{ x } ^ { 2 } - 4 x - 5 = 0
x
2
−
4
x
−
5
=
0
三角法
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
θ
cos
θ
=
2
sin
θ
一次方程式
y = 3x + 4
y
=
3
x
+
4
算術
699 * 533
6
9
9
∗
5
3
3
マトリックス
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
[
2
5
3
4
]
[
2
−
1
0
1
3
5
]
連立方程式
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
{
8
x
+
2
y
=
4
6
7
x
+
3
y
=
4
7
微分法
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
d
x
d
(
x
−
5
)
(
3
x
2
−
2
)
積分法
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
∫
0
1
x
e
−
x
2
d
x
限界
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
x
→
−
3
lim
x
2
+
2
x
−
3
x
2
−
9
トップに戻る