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Chapter 3: Problem 33

Prove the formula $$ \cos ^{2} x=\frac{1+\cos 2 x}{2} $$

Short Answer

Expert verified
Recall the double-angle formula for cosine, \(\cos 2x = \cos^2 x - \sin^2 x\). Express \(\sin^2 x\) in terms of \(\cos^2 x\) using the identity \(\sin^2 x + \cos^2 x = 1\), so \(\sin^2 x = 1 - \cos^2 x\). Substitute this expression in the double-angle formula: \(\cos 2x = \cos^2 x - (1 - \cos^2 x)\). Simplify the expression, \(\cos 2x = 2\cos^2 x - 1\), and solve for \(\cos^2 x\): \(\cos^2 x = \frac{1 + \cos 2x}{2}\). This confirms the given formula is true: \(\cos^{2}x = \frac{1+\cos 2x}{2}\).

Step by step solution

01

Recall double-angle formula for cosine

Recall the double-angle formula for cosine: \(\cos 2x = \cos^2 x - \sin^2 x\)
02

Express sine squared in terms of cosine squared

We know that \(\sin^2 x + \cos^2 x = 1\). From this, we can isolate \(\sin^2 x\) and express it in terms of \(\cos^2 x\): \(\sin^2 x = 1 - \cos^2 x\)
03

Substitute sine squared in the double-angle formula

Substitute the expression for \(\sin^2 x\) in the double-angle formula to get an expression only involving \(\cos x\): \(\cos 2x = \cos^2 x - (1 - \cos^2 x)\)
04

Simplify the expression

Simplify the expression to get the expression for \(\cos^2 x\): \(\cos 2x = \cos^2 x - 1 + \cos^2 x\) \(\cos 2x = 2\cos^2 x - 1\)
05

Solve for cosine squared

Finally, solve for \(\cos^2 x\): \(\cos^2 x = \frac{1 + \cos 2x}{2}\) This confirms that the given formula is true: \(\cos ^{2} x=\frac{1+\cos 2 x}{2}\).

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Key Concepts

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

Double-Angle Formulas
The double-angle formulas in trigonometry are powerful tools that help simplify expressions involving trigonometric functions. These formulas can express the trigonometric functions of double angles in terms of single angles.

For the cosine function, the double-angle formula is given by:
  • \(\cos 2x = \cos^2 x - \sin^2 x\)
This formula can also be rewritten using only the cosine function by applying the Pythagorean identity, which allows substitution of all sine terms in terms of cosine:
  • \(\cos 2x = 2\cos^2 x - 1\)
Double-angle formulas make calculations easier, especially when needing to expand or simplify expressions where multiple angles are involved. Using them can simplify complex trigonometric identities and integrals into more manageable terms.
Cosine Function
The cosine function is one of the primary trigonometric functions. It measures the length of the adjacent side compared to the hypotenuse in a right-angled triangle. As one of the fundamental trigonometric functions, cosine has key properties including periodicity, symmetry, and specific values at notable angles.

The cosine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians.
  • The graph of the cosine function is a wave starting at 1 when \(x = 0\) and shows symmetry around the y-axis.
  • Important values include \(\cos(0) = 1\), \(\cos(\pi/2) = 0\), and \(\cos(\pi) = -1\).
Understanding the cosine function is crucial for working with more advanced trigonometric identities, such as those encountered in calculus and wave motion.
Pythagorean Identity
Pythagorean identities are essential relationships in trigonometry that come from the Pythagorean theorem applied in the unit circle. The most basic of these identities relates the square of sine and cosine:
  • \(\sin^2 x + \cos^2 x = 1\)
This identity is fundamental in expressing one trigonometric function in terms of another. For instance, you can express \(\sin^2 x\) using safe algebraic manipulation:
  • \(\sin^2 x = 1 - \cos^2 x\)
This rearrangement is very useful when solving trigonometric equations or proving identities, such as double-angle formulas where both sine and cosine need to be incorporated effectively. Understanding how to manipulate and apply the Pythagorean identity is vital for tackling more complex trigonometric proofs and derivations.

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Most popular questions from this chapter

Use Newton's method to find the point of intersection of the graphs to four decimal places of accuracy by solving the equation \(f(x)-g(x)=0 .\) Use the initial estimate \(x_{0}\) for the \(x\) -coordinate. f(x)=\tan x, g(x)=1-x, \quad x_{0}=1

Maximum Power Output Suppose that the source of current in an electric circuit is a battery. Then the power output \(P\) (in watts) obtained if the circuit has a resistance of \(R\) ohms is given by $$ P=\frac{E^{2} R}{(R+r)^{2}} $$ where \(E\) is the electromotive force in volts and \(r\) is the internal resistance of the battery in ohms. If \(E\) and \(r\) are constant, find the value of \(R\) that will result in the greatest power output. What is the maximum power output?

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Estimate the value of the radical accurate to four decimal places by using three iterations of Newton's method to solve the equation \(f(x)=0\) with initial estimate \(x_{0}\). \(\sqrt[4]{20} ; \quad f(x)=x^{4}-20 ; x_{0}=2.1\)

Use Newton's method to approximate the indicated zero of the function. Continue with the iteration until two successive approximations differ by less than \(0.0001\). The zero of \(f(x)=x^{5}+x-1\) between \(x=0\) and \(x=1\). Take \(x_{0}=0.5\).
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