17Calculus Precalculus - Matrix Inverse
This page covers matrix inverses. One big application of matrix inverses is solving linear systems using an inverse matrix.
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If you are given a matrix \(A\), the inverse of \(A\) is defined as the matrix, let's call it \(B\) for now, where \(AB=BA=I\). This is similar to the inverse of a number. For example, the inverse of \(2\) is \(1/2\) since \(2(1/2)=1\). Sometimes we even write \(1/2=2^{-1}\). So \(2(2^{-1})=1\).
The same idea holds for matrices. So we call the inverse of matrix \(A\), \(A^{-1}\). This gives us \(AA^{-1}=1\) and \(A^{-1}A=1\). Notice that this is a special case of when matrix multiplication is commutative. The point of this page is learning how to calculate the inverse matrix \(A^{-1}\).
Determining the inverse of a matrix is, in general, not an easy task. And, as you know from algebra, an inverse may not exist. For example, there is no inverse of zero. So how can you tell when a matrix doesn't have an inverse? Well, there is a simple test you can do before you start calculating the inverse to see if one exists. Calculate the determinant. If the determinant is zero, then the inverse doesn't exist.
Okay, so if you have calculated the determinant and found it to be non-zero, then you can find an inverse matrix. There are several techniques, some easier than others, including using a formula (for 2x2), row reduction and using minors and cofactors.
Formula for 2x2 Matrices
For 2x2 matrices there is a formula that you can use. (As always, check with your instructor to see if they allow you to use this special case technique.)
\(\displaystyle{ A = \left[ \begin{array}{rr} a & b \\ c & d \end{array}\right] }\) |
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\(\displaystyle{ A^{-1} = \frac{1}{ad-bc} \left[ \begin{array}{rr} d & -b \\ -c & a \end{array} \right]}\) |
Let's look at this closely. Notice that the fraction out in front has the determinant of matrix \(A\) in the denominator. The matrix part of the inverse can be summed up in these two rules.
1. Swap the upper-left and lower-right terms.
2. Negate the other two terms but leave them in the same positions.
Important Note - Be careful to use this only on 2x2 matrices. This will not work on 3x3 or any other size of matrix.
Here is a great video explaining this in more detail.
Dr Chris Tisdell - Inverse of Matrix 2 x 2 Case [10min-24secs]
video by Dr Chris Tisdell |
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Before you go on, work these 2x2 practice problems to make sure you can easily use this technique.
Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 7 & 2 \\ 17 & 5 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 7 & 2 \\ 17 & 5 \end{array}\right]}\).
Solution |
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In this solution, he shows how to confirm that the answer is the inverse of the original matrix by calculating \(A A^{-1}\) and making sure it is equal to \(I_2\). This is only a partial confirmation. He would need to calculate \(A^{-1} A\) to confirm that it equals \(I_2\) as well. This is always good to do, especially on exams when you have time.
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 8 & 6 \\ 5 & 4 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 8 & 6 \\ 5 & 4 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 3 & 5 \\ 1 & -2 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 3 & 5 \\ 1 & -2 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 2 & 4 \\ 1 & 3 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use the 2x2 formula to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 2 & 4 \\ 1 & 3 \end{array}\right]}\).
Solution |
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Row Reduction
It would be great if there was a formula to calculate the inverse of any matrix as easily as the 2x2 case. But, alas, such is not the case. One common technique to find the inverse of any matrix is to use row reduction. The technique uses the same operations as Gaussian Elimination/Row Reduction. This next video shows how to do that with an example.
Dr Chris Tisdell - How to compute the inverse of a 3 x 3 matrix [14min-12secs]
video by Dr Chris Tisdell |
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Before we go on, let's make sure you have a good understanding of this idea. Use row reduction to find the inverses of these matrices.
Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array}\right]}\). Check your answer by using the 2x2 formula.
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array}\right]}\). Check your answer by using the 2x2 formula.
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 3 & 2 \\ 4 & 3 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rr} 3 & 2 \\ 4 & 3 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{J = \left[\begin{array}{rr} 1 & 3 \\ 2 & 5\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{J = \left[\begin{array}{rr} 1 & 3 \\ 2 & 5\end{array}\right]}\).
Final Answer |
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\(\displaystyle{J^{-1} = \left[\begin{array}{rr} -5 & 3 \\ 2 & -1\end{array}\right]}\)
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{J = \left[\begin{array}{rr} 1 & 3 \\ 2 & 5\end{array}\right]}\).
Solution |
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Final Answer |
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\(\displaystyle{J^{-1} = \left[\begin{array}{rr} -5 & 3 \\ 2 & -1\end{array}\right]}\) |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{M = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{M = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2\end{array}\right]}\).
Final Answer |
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\(\displaystyle{M^{-1} = \left[\begin{array}{rrr} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1\end{array}\right]}\)
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{M = \left[\begin{array}{rrr} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2\end{array}\right]}\).
Solution |
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2024 video
video by Michel vanBiezen |
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Final Answer |
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\(\displaystyle{M^{-1} = \left[\begin{array}{rrr} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1\end{array}\right]}\) |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 1 & 0 & 2 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 1 & 0 & 2 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]}\).
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 4 & 1 \\ -1 & 1 & -1 \\ 1 & 4 & 0\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 4 & 1 \\ -1 & 1 & -1 \\ 1 & 4 & 0\end{array}\right]}\).
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 3 & 0 \\ 1 & -2 & 1 \\ 2 & 0 & -1 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 3 & 0 \\ 1 & -2 & 1 \\ 2 & 0 & -1 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{array}\right]}\).
Solution |
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2776 video
video by Michel vanBiezen |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & 2 \\ 1 & 2 & -1 \\ -1 & 1 & 4 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use row reduction to find the inverse of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & 2 \\ 1 & 2 & -1 \\ -1 & 1 & 4 \end{array}\right]}\).
Solution |
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Determinants and Cofactors |
The most involved technique uses determinants and cofactors to calculate an inverse. This will work with any size of matrix and is therefore the most general. Here is a video that shows how to do this with an example.
Khan Academy - Calculating matrix of minors and cofactor matrix [8min-46secs]
video by Khan Academy |
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Use determinants and cofactors to find the inverses of these matrices.
Unless otherwise instructed, use determinants and cofactors to find the inverse of the matrix \(\displaystyle{K = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 2 & -1 & 3 \\ 1 & 1 & 4\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverse of the matrix \(\displaystyle{K = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 2 & -1 & 3 \\ 1 & 1 & 4\end{array}\right]}\).
Final Answer |
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\(\displaystyle{K^{-1}=\frac{1}{3}\left[\begin{array}{rrr} -7 & 1 & 1 \\ -5 & -1 & 2 \\ 3 & 0 & 0\end{array}\right]}\)
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverse of the matrix \(\displaystyle{K = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 2 & -1 & 3 \\ 1 & 1 & 4\end{array}\right]}\).
Solution |
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2022 video
video by PatrickJMT |
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Final Answer |
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\(\displaystyle{K^{-1}=\frac{1}{3}\left[\begin{array}{rrr} -7 & 1 & 1 \\ -5 & -1 & 2 \\ 3 & 0 & 0\end{array}\right]}\) |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{L = \left[\begin{array}{rrr} 3 & -1 & 2 \\ 5 & 1 & 0 \\ -2 & 3 & 4\end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{L = \left[\begin{array}{rrr} 3 & -1 & 2 \\ 5 & 1 & 0 \\ -2 & 3 & 4\end{array}\right]}\).
Final Answer |
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\(\displaystyle{L^{-1} = \frac{1}{66}\left[\begin{array}{rrr} 4 & 10 & -2 \\ -20 & 16 & 10 \\ 17 & -7 & 8\end{array}\right]}\)
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{L = \left[\begin{array}{rrr} 3 & -1 & 2 \\ 5 & 1 & 0 \\ -2 & 3 & 4\end{array}\right]}\).
Solution |
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2023 video
video by PatrickJMT |
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Final Answer |
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\(\displaystyle{L^{-1} = \frac{1}{66}\left[\begin{array}{rrr} 4 & 10 & -2 \\ -20 & 16 & 10 \\ 17 & -7 & 8\end{array}\right]}\) |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & 2 \\ 3 & -2 & 1 \\ 2 & 1 & -1 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 2 & 2 \\ 3 & -2 & 1 \\ 2 & 1 & -1 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & -2 & -1 \\ 3 & 0 & 0 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & -2 & -1 \\ 3 & 0 & 0 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 1 & 0 \\ 1 & -1 & 1 \\ 3 & 2 & 1 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 2 & 1 & 0 \\ 1 & -1 & 1 \\ 3 & 2 & 1 \end{array}\right]}\).
Solution |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 3 & 0 & 2 \\ 2 & 0 & -2 \\ 0 & 1 & 1 \end{array}\right]}\).
Problem Statement |
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Unless otherwise instructed, use determinants and cofactors to find the inverses of the matrix \(\displaystyle{ \left[\begin{array}{rrr} 3 & 0 & 2 \\ 2 & 0 & -2 \\ 0 & 1 & 1 \end{array}\right]}\).
Solution |
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Okay, so now that you know how to find the inverse of a matrix, you can use it to solve systems of linear equations.
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Really UNDERSTAND Precalculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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