PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Trigonometric Identities
Trigonometry in mathematics deals with the right-angled triangle’s angles and sides. By trigonometric identities, we mean the identities we use whenever we need to express the various trigonometric functions in terms of an equation.
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse of normal trigonometric functions. Alternatively denoted as cyclometric or arcus functions, these inverse trigonometric functions exist to counter the basic trigonometric functions, such as sine (sin), cosi…
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Question
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Chapter 5.4, Problem 16E
To determine

To find: the exact values of the sine, cosine and tangent of the angle.

Expert Solution & Answer
Check Mark

Answer to Problem 16E

. sin(255°)=−(6+24) , cos255°=(2−64) , tan255°=2−3

Explanation of Solution

Given information:

Given angle,

  255°=300°−45° .

Formula used:

  sin(u−v)=sinucosv−cosusinvcos(u−v)=cosucosv+sinusinvtan(u−v)=tanu−tanv1+tanutanv

Calculation:

Find the value of sine, cosine and tangent of the angle 255° by using the difference formulas as follows

  sin255°=sin(300°−45°)=sin300°cos45°−cos300°sin45°=(−32)(12)−(12)(12)=−322−(122)sin255°=−(322+122)=−(3+122)

Multiplying numerator and denominator by 2

  sin(255°)=(3+1)2(22)2=−(6+24)

Hence,

  sin(255°)=−(6+24) .

Find exact values of the cosine of the above angle as below

Taking cosine on both sides as below

  cos255°=cos(300°−45°)

Using the sum formula cos(u−v)=cosucosv+sinusinv , we get

  cos255°=cos(300°−45°)=cos300°cos45°+sin300°sin45°=(12)(12)+(−32)(−12)=122−(322)cos255°=(1−322)

Multiplying numerator and denominator by 2

  cos255°=((1−3)2(22)2)=(2−64)

Hence,

  cos255°=(2−64) .

Find the exact values of the tangent of the above angle as below

Taking tangent on both sides as below

  tan255°=tan(300°−45°)=tan300°−tan45°1+tan300°tan45°=(−3)−11+(−3)×1=−(1+3)(1−3)

Hence,

  tan255°=−(1+3)(1−3)=(1−3)21−3=4−232=−3+2 .

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Chapter 5.4, Problem 15E
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Chapter 5.4, Problem 17E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 5.1 - Prob. 1ECh. 5.1 - Prob. 2ECh. 5.1 - Prob. 3ECh. 5.1 - Prob. 4ECh. 5.1 - Prob. 5ECh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
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