/*! 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 18 Find the exact value of each exp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the exact value of each expression. $$\cos \left(240^{\circ}+45^{\circ}\right)$$

Short Answer

Expert verified
\(\cos(240^{\circ}+45^{\circ}) = -((- \sqrt{2} + \sqrt{6})/ 4\)

Step by step solution

01

Conversion

First, convert the sum of angles expressed in degrees into a single angle, by addition. Therefore, \(240^{\circ}+45^{\circ}= 285^{\circ}\)
02

Reference Angle

Identify the equivalent reference angle in the first quadrant, knowing the angle is in the third quadrant. So the reference angle is \(285^{\circ}-180^{\circ}= 105^{\circ}\)
03

Calculate Cosine

Find the cosine of the reference angle, knowing that cosine of an angle in the third quadrant is negative. Hence, \(\cos(285^{\circ}) = -\cos(105^{\circ})\)
04

Breakdown of Angle

Breakdown the \(105^{\circ}\) into sum of two known angles: \(105^{\circ}= 60^{\circ}+ 45^{\circ}\)
05

Use Cosine of Sum of Angles Rule

Apply the rule of cosine of sum of two angles on \(60^{\circ}+ 45^{\circ}\), which is \(\cos(A + B) = \cos A \cos B - \sin A \sin B\). So \(\cos(105^{\circ}) = -(\cos(60^{\circ})\cos(45^{\circ}) - \sin(60^{\circ})\sin(45^{\circ}))\)
06

Calculation

Perform the calculation, substituting known values from special angles \(\cos(60^{\circ}) = 1/2\), \(\cos(45^{\circ}) = \sin(45^{\circ}) = \sin(60^{\circ}) = \sqrt{2}/2\). Hence, \(\cos(285^{\circ}) = -((1/2) * (\sqrt{2}/2) - (\sqrt{2}/2) * (\sqrt{3}/2) ) = -((- \sqrt{2} + \sqrt{6})/ 4\)

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

Most popular questions from this chapter

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$\tan x=-4.7143$$

Use this information to solve. When throwing an object, the distance achieved depends on its initial velocity, \(v_{0}\) and the angle above the horizontal at which the object is thrown, \(\theta\) The distance, \(d\), in feet, that describes the range covered is given by $$d=\frac{v_{0}^{2}}{16} \sin \theta \cos \theta$$ where \(v_{0}\) is measured in feet per second. You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of \(v_{0}=90\) feet per second, at what angle of elevation, \(\theta,\) to the nearest degree, should you direct your throw so that it can be easily caught by your friend located 170 feet away?

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$5 \cot ^{2} x-15=0$$

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$2 \sin ^{2} x=3-\sin x$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning.I've noticed that for sine, cosine, and tangent, the trig function for the sum of two angles is not equal to that trig function of the first angle plus that trig function of the second angle.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.