/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Evaluate (if possible) the funct... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(f(x)=\cos 2 x\) (a) \(f(0)\) (b) \(f(-\pi / 4)\) (c) \(f(\pi / 3)\)

Short Answer

Expert verified
The evaluations of the function at the given values are: (a) \(f(0) = 1\), (b) \(f(-\pi / 4) = 0\), and (c) \(f(\pi / 3) = -1/2\).

Step by step solution

01

Evaluate (a) f(0)

For \(f(0)\), substitutute \(x = 0\) in the function \(f(x)=\cos 2x\), which gives \(f(0) = \cos 2(0) = \cos (0)\). The function value of cos at 0 is 1, so \(f(0) = 1\).
02

Evaluate (b) f(-\(\pi / 4\))

For \(f(-\pi / 4)\), substituting \(x = -\pi / 4\) in the function \(f(x)=\cos 2x\) gives \(f(-\pi / 4) = \cos 2(-\pi/4) = \cos (-\pi/2)\). Now, \(cos (-\theta)\) is equal to \(cos (\theta)\), thus \(cos (-\pi / 2)\) = 0. So, \(f(-\pi / 4) = 0\).
03

Evaluate (c) f(\(\pi / 3\))

For \(f(\pi / 3)\), substituting \(x = \pi / 3\) in the function \(f(x)=\cos 2x\) gives \(f(\pi / 3) = \cos 2(\pi/3) = \cos (2\pi/3)\). This is a commonly known value, which equals to -1/2. So, \(f(\pi / 3) = -1/2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.