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Chapter 15: Problem 7

(a) From the addition formula for sin and cos derive formulas for \(\sin 2 x\) \(\cos 2 x, \sin 3 x,\) and \(\cos 3 x\) (b) Use these formulas to find the following values of the trigonometric functions (usually deduced by geometric arguments in elementary trigonometry): $$ \begin{aligned} \sin \frac{\pi}{4} &=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ \tan \frac{\pi}{4} &=1 \\ \sin \frac{\pi}{6} &=\frac{1}{2} \\ \cos \frac{\pi}{6} &=\frac{\sqrt{3}}{2} \end{aligned} $$

Short Answer

Expert verified
Derived: \sin 2x = 2 \sin x \cos x, \cos 2x = \cos^2 x - \sin^2 x, \sin 3x = 3 \sin x - 4 \sin^3 x, and \cos 3x = 4 \cos^3 x - 3 \cos x. Verified given trigonometric values.

Step by step solution

01

Derive the formula for \( \sin 2x \)

Using the sine addition formula \[ \sin(a + b) = \sin a \cos b + \cos a \sin b\], set \ a = b = x\:\ \[ \sin 2x = \sin(x + x) = \sin x \cos x + \cos x \sin x = 2 \sin x \cos x\]
02

Derive the formula for \(\cos 2x\)

Using the cosine addition formula \[ \cos(a + b) = \cos a \cos b - \sin a \sin b\], set \a = b = x\: \[\cos 2x = \cos(x + x) = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x\]\Alternatively, using \[\cos^2 x + \sin^2 x = 1\]\, we get: \[\cos 2x = 2 \cos^2 x - 1\]\and \[\cos 2x = 1 - 2 \sin^2 x \]
03

Derive the formula for \(\sin 3x\)

Using the angle addition formula for sine multiple times, break down the expression: \[\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x\]\Substitute \sin 2x and \cos 2x with the formulas derived earlier: \[ \sin 3x = (2 \sin x \cos x) \cos x + (\cos^2 x - \sin^2 x) \sin x\]\Simplify it to: \[ \sin 3x = 3 \sin x - 4 \sin^3 x\]
04

Derive the formula for \(\cos 3x\)

Similarly, using the angle addition formula for cosine and breaking down the expression: \[\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\]\Substitute \sin 2x and \cos 2x with the formulas derived earlier: \[ \cos 3x = (\cos^2 x - \sin^2 x) \cos x - (2 \sin x \cos x) \sin x\]\Simplify it to: \[ \cos 3x = 4 \cos^3 x - 3 \cos x\]
05

Verify trigonometric values using derived formulas

Verify the given trigonometric values using the derived formulas for simpler angles or previously known values: \Given: \[ \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\]\, \[ \tan \frac{\pi}{4} = 1 \]\[ \sin \frac{\pi}{6} = \frac{1}{2}\]\[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \]

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Key Concepts

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

Angle Addition Formulas
The angle addition formulas are essential in trigonometry. They allow you to find the sine or cosine of the sum of two angles. For sine, the formula is \(\sin(a + b) = \sin a \cos b + \cos a \sin b\). For cosine, it's \(\cos(a + b) = \cos a \cos b - \sin a \sin b\).
By using these formulas, you can derive other useful formulas for multiple angles, like \(\sin 2x\) and \(\cos 2x\).
To derive \(\sin 2x\), let \[ a = b = x\]. Thus, \(\sin 2x = \sin(x + x) = \sin x \cos x + \cos x \sin x = 2 \sin x \cos x\).
Similarly, for \(\cos 2x\), also let \[ a = b = x\]. Hence, \(\cos 2x = \cos(x + x) = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x\).
These fundamental formulas are stepping stones to derive other complex trigonometric identities.
Deriving Trigonometric Values
Working out specific values for trigonometric functions often involves these angle addition formulas. For example, to derive \(\sin \frac{\pi}{4}\) and \(\cos \frac{\pi}{4}\), we use known values from the unit circle and symmetries in trigonometric properties.
From geometry, we know:
  • \(\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\)
  • \(\tan \frac{\pi}{4} = 1\)
  • \(\sin \frac{\pi}{6} = \frac{1}{2}\)
  • \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\)

Using these known values can help verify results derived using the addition formulas. By substituting into derived identities like \(\sin 2x\) and \(\cos 3x\), they affirm the correctness of complex formulas.
Double and Triple Angle Formulas
Double and triple angle formulas help simplify the process of finding trigonometric values for multiple angles.
For double angles, \(\sin 2x\) and \(\cos 2x\) are derived using the angle addition formulas.
For \(\sin 2x\), we get: \(\sin 2x = 2 \sin x \cos x\). \(\cos 2x\) can be written as \(\cos 2x = 2 \cos^2 x - 1\) or \(1 - 2 \sin^2 x\).
For triple angles, split \(3x\) into \(2x + x\) and use the angle addition formulas repeatedly. For example:
  • \sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x\
  • Substitute \(\sin 2x\) and \(\cos 2x\) to get,\sin 3x = 3 \sin x - 4 \sin^3 x.
  • For \(\cos 3x\), \(\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\).
  • Substituting, we get \(\cos 3x = 4 \cos^3 x - 3 \cos x\).

Understand these step-by-step and use them to simplify complex expressions easily.

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Most popular questions from this chapter

Prove that if \(m\) and \(n\) are any numbers, then $$ \begin{array}{l} \sin m x \sin n x=\frac{1}{2}[\cos (m-n) x-\cos (m+n) x] \\ \sin m x \cos n x=\frac{1}{2}[\sin (m+n) x+\sin (m-n) x] \\ \cos m x \cos n x=\frac{1}{2}[\cos (m+n) x+\cos (m-n) x] \end{array} $$

Graph the following functions. (a) \(f(x)=\sin 2 x\) (b) \(f(x)=\sin \left(x^{2}\right) .\) (A pretty respectable sketch of this graph can be obtained using only a picture of the graph of sin. Indeed, pure thought is your only hope in this problem, because determining the sign of the derivative \(f^{\prime}(x)=\cos \left(x^{2}\right) \cdot 2 x\) is no easier than determining the behavior of \(f\) directly. The formula for \(f^{\prime}(x)\) does indicate one important fact, however \(-f^{\prime}(0)=0,\) which must be true since \(j\) is even, and which should be clear in your graph.) (c) \(f(x)=\sin x+\sin 2 x .\) (It will probably be instructive to first draw the graphs of \(g(x)=\sin x\) and \(h(x)=\sin 2 x\) carefully on the same set of axes, from 0 to \(2 \pi,\) and guess what the sum will look like. You can easily find out how many critical points \(f\) has on \([0,2 \pi]\) by considering the derivative of \(f .\) You can then determine the nature of these critical points by finding out the sign of \(f\) at each point; your sketch will probably suggest the answer.) (d) \(f(x)=\tan x-x .\) (First determine the behavior of \(f\) in \((-\pi / 2, \pi / 2)\) in the intervals \((k \pi-\pi / 2, k \pi+\pi / 2)\) the graph of \(f\) will look exactly the same, except moved up a certain amount. Why?) (e) \(f(x)=\sin x-x .\) (The material in the Appendix to Chapter 11 will be particularly helpful for this function.) (f) \(f(x)=\left\\{\begin{array}{ll}\frac{\sin x}{x}, & x \neq 0 \\ 1, & x=0\end{array}\right.\) (Part (d) should enable you to determine approximately where the zeros of \(f^{\prime}\) are located. Notice that \(f\) is even and continuous at \(0 ;\) also consider the size of \(f\) for large \(x .\) ) (g) \(f(x)=x \sin x\)

(a) If \(f\) is integrable on \([-\pi, \pi],\) show that the minimum value of occurs when $$ \int_{-\pi}^{\pi}(f(x)-a \cos n x)^{2} d x $$ $$ a=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x $$ and the minimum value of $$ \int_{-\pi}^{\pi}(f(x)-a \sin n x)^{2} d x $$ when $$ a=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x $$ (In each case, bring a outside the integral sign, obtaining a quadratic expression in \(a .)\) (b) Define $$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x, \quad n=0,1,2, \ldots \\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x, \quad n=1,2,3, \ldots \end{aligned} $$ Show that if \(c_{i}\) and \(d_{i}\) are any numbers, then \(\int_{-\pi}^{\pi}\left(f(x)-\left[\frac{c_{0}}{2}+\sum_{n=1}^{N} c_{n} \cos n x+d_{n} \sin n x\right]\right)^{2} d x\) \(\quad=\int_{-\pi}^{\pi}[f(x)]^{2} d x-2 \pi\left(\frac{a_{0} c_{0}}{2}+\sum_{n=1}^{N} a_{n} c_{n}+b_{n} d_{n}\right)+\pi\left(\frac{c_{0}^{2}}{2}+\sum_{n=1}^{N} c_{n}^{2}+d_{n}^{2}\right)\) \(=\int_{-\pi}^{\pi}[f(x)]^{2} d x-\pi\left(\frac{a_{0}^{2}}{2}+\sum_{n=1}^{N} a_{n}^{2}+b_{n}^{2}\right)\) \(\quad+\pi\left(\left(\frac{c_{0}}{\sqrt{2}}-\frac{a_{0}}{\sqrt{2}}\right)^{2}+\sum_{n=1}^{N}\left(c_{n}-a_{n}\right)^{2}+\left(d_{n}-b_{n}\right)^{2}\right)\) thus showing that the first integral is smallest when \(a_{i}=c_{i}\) and \(b_{i}=d_{i} .\) In other words, among all "linear combinations" of the functions \(s_{n}(x)=\sin n x\) and \(c_{n}(x)=\cos n x\) for \(1 \leq n \leq N,\) the particular function \(g(x)=\frac{a_{0}}{2}+\sum_{n=1}^{N} a_{n} \cos n x+b_{n} \sin n x\) has the "closest fit" to \(f\) on \([-\pi, \pi]\)

(a) After all the work involved in the definition of sin, it would be disconcerting to find that sin is actually a rational function. Prove tha: it isn't. (There is a simple property of sin which a rational functio: cannot possible have.) (b) Prove that sin isn't even defined implicitly by an algebraic equation. that is, there do not exist rational functions \(f_{0}, \ldots, f_{n-1}\) such that \((\sin x)^{n}+f_{n-1}(x)(\sin x)^{n-1}+\cdots+f_{0}(x)=0 \quad\) for all \(x\) Hint: Prove that \(f_{0}=0,\) so that \(\sin x\) can be factored out. The remaining factor is 0 except perhaps at multiples of \(2 \pi .\) But this implies that it is 0 for all \(x .\) (Why?) You are now set up for a proo: by induction.

Let \(f(x)=\left\\{\begin{array}{ll}\frac{\sin x}{x}, & x \neq 0 \\ 1, & x=0\end{array}\right.\) (a) Find \(f^{\prime}(0)\) (b) Find \(f^{\prime \prime}(0)\) At this point, you will almost certainly have to use l'H贸pital's Rule, but in Chapter 23 we will be able to find \(f^{(k)}(0)\) for all \(k,\) with almost no work at all.
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