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In elementary trigonometry, you probably learned the law of cosines for a triangle of sides \(a, b,\) and \(c,\) that \(c^{2}=a^{2}+b^{2}-2 a b \cos \theta,\) where \(\theta\) is the angle between the sides \(a\) and \(b\). Show that the law of cosines is an immediate consequence of the identity \((\mathbf{a}+\mathbf{b})^{2}=a^{2}+b^{2}+2 \mathbf{a} \cdot \mathbf{b}\)

Short Answer

Expert verified
By expanding the identity and substituting the dot product formula, the law of cosines is verified.

Step by step solution

01

Expanding the Given Identity

The identity given is \((\mathbf{a} + \mathbf{b})^2 = a^2 + b^2 + 2 \mathbf{a} \cdot \mathbf{b}\).This identity holds for any vectors \(\mathbf{a}\) and \(\mathbf{b}\). Expanding the left side results in\((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}\),which confirms the identity with \(a^2 = \mathbf{a} \cdot \mathbf{a}\) and \(b^2 = \mathbf{b} \cdot \mathbf{b}\).
02

Relating Vectors to Triangle Sides

Consider a triangle where the sides are represented as vectors. Let \(\mathbf{c} = \mathbf{a} + \mathbf{b}\) be the vector from one vertex to another. This means that the square of the magnitude of \(\mathbf{c}\) is\[ c^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}). \]Substituting the identity from Step 1,\[ c^2 = a^2 + b^2 + 2 \mathbf{a} \cdot \mathbf{b}. \]
03

Incorporating the Angle Between Vectors

Using the dot product relationship, we know that\( \mathbf{a} \cdot \mathbf{b} = a b \cos \theta \),where \(\theta\) is the angle between the vectors \(\mathbf{a}\) and \(\mathbf{b}\).Substituting into the equation from Step 2, we get\[ c^2 = a^2 + b^2 + 2 a b \cos \theta. \]
04

Deriving the Law of Cosines Form

Manipulate the equation from Step 3 to reach the form given in the law of cosines:\[ c^2 = a^2 + b^2 + 2 a b \cos(\theta), \]which can be rewritten as\[ c^2 = a^2 + b^2 - 2 a b \cos(180° - \theta). \]Since \(\cos(180° - \theta) = -\cos(\theta)\), this simplifies back to\[ c^2 = a^2 + b^2 - 2 a b \cos \theta. \]The law of cosines is thus confirmed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry
Trigonometry is the branch of mathematics that deals with the study of triangles, particularly right triangles, and the relationships between their angles and sides. One of the most powerful tools in trigonometry is the Law of Cosines, which helps to find unknown sides or angles in any triangle, not just right-angled ones. The formula for the Law of Cosines is: \[c^2 = a^2 + b^2 - 2ab \cos(\theta).\]This formula shows how the sides of a triangle relate through angles. When you know two sides of a triangle and the angle between them, the Law of Cosines allows you to solve for the third side. This is especially useful when the triangle is not a right triangle, where simpler tools like the Pythagorean theorem would not apply. Understanding the Law of Cosines involves knowing how angles affect the length of sides across triangles. Familiarize yourself with concepts of angles measured in radians or degrees to fully utilize this law in solving problems.
Vector Algebra
Vector Algebra is a way to understand and solve geometric problems using vectors, which are quantities that have both magnitude and direction. In the context of the Law of Cosines, vectors can represent the sides of a triangle. Think of each side of a triangle as a vector. If vectors \( \mathbf{a} \) and \( \mathbf{b} \) represent two sides, then the vector \( \mathbf{c} \) can represent the resultant side, calculated using vector addition, such as \( \mathbf{c} = \mathbf{a} + \mathbf{b} \).Vectors make it possible to calculate the magnitude and direction of a resultant vector using principles of vector addition and subtraction, which directly ties back to the Law of Cosines. Vector Algebra involves operations like addition, subtraction, and scalar multiplication, all of which lay the foundation to use vectors in describing sides of triangles. The intersection of these vectors at particular angles will provide different magnitudes for the results, hence connecting to how the Law of Cosines predicts the sides of triangles. Knowing how vectors behave, through vector addition or orientation, gives a powerful way to compute side lengths of triangles beyond basic geometry.
Dot Product
The Dot Product, also known as the scalar product, is a critical tool in understanding the relationship between vectors. It computes a single number from two vectors, providing insights into the angle between them. Mathematically, the dot product is defined as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \), where \( \theta \) is the angle between the two vectors.In terms of the Law of Cosines, the dot product connects the vectors representing the sides of a triangle. By substituting \( \mathbf{a} \cdot \mathbf{b} = ab \cos(\theta) \) into the equation for \( c^2 \), we derive how dots on the sides reflect the angle's influence between them. This makes the dot product an invaluable part of vector calculation, helping us relate vector algebra to geometric concepts like the angles in a triangle.The dot product helps us understand projection: how one vector projects onto another, which translates to how one side of the triangle influences another based on the angle. This deepens comprehension of how the Law of Cosines bridges between abstract vector concepts and concrete trigonometric applications. Understanding dot products enriches problem-solving by offering insights into how triangles' sides and angles interrelate through vector properties.

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Most popular questions from this chapter

The unknown vector \(\mathbf{v}\) satisfies \(\mathbf{b} \cdot \mathbf{v}=\lambda\) and \(\mathbf{b} \times \mathbf{v}=\mathbf{c},\) where \(\lambda, \mathbf{b},\) and \(\mathbf{c}\) are fixed and known. Find \(\mathbf{v}\) in terms of \(\lambda, \mathbf{b},\) and \(\mathbf{c}\)

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure \(1.8 .\) The electric and magnetic fields at a point \(\mathbf{r}_{1}\) due to a charge \(q_{2}\) at \(\mathbf{r}_{2}\) moving with constant velocity \(\mathbf{v}_{2}\) (with \(v_{2} \ll c\) ) are \(^{15}\) $$\mathbf{E}\left(\mathbf{r}_{1}\right)=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \quad \text { and } \quad \mathbf{B}\left(\mathbf{r}_{1}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}}$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2}\) is the vector pointing from \(\mathbf{r}_{2}\) to \(\mathbf{r}_{1}\). (The first of these you should recognize as Coulomb's law.) If \(\mathbf{F}_{12}^{\mathrm{el}}\) and \(\mathbf{F}_{12}^{\text {mag }}\) denote the electric and magnetic forces on a charge \(q_{1}\) at \(\mathbf{r}_{1}\) with velocity \(\mathbf{v}_{1},\) show that \(F_{12}^{\operatorname{mag}} \leq\left(v_{1} v_{2} / c^{2}\right) F_{12}^{\mathrm{el}} .\) This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

A ball is thrown with initial speed \(v_{\mathrm{o}}\) up an inclined plane. The plane is inclined at an angle \(\phi\) above the horizontal, and the ball's initial velocity is at an angle \(\theta\) above the plane. Choose axes with \(x\) measured up the slope, \(y\) normal to the slope, and \(z\) across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance \(R=2 v_{\mathrm{o}}^{2} \sin \theta \cos (\theta+\phi) /\left(g \cos ^{2} \phi\right)\) from its launch point. Show that for given \(v_{\mathrm{o}}\) and \(\phi,\) the maximum possible range up the inclined plane is \(R_{\max }=v_{\mathrm{o}}^{2} /[g(1+\sin \phi)]\)

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