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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Chapter 5.5, Problem 52E
To determine

Tofind:The exact values of the sine, cosine and tangent at angle 67°30′ using half-angle formula.

Expert Solution & Answer
Check Mark

Answer to Problem 52E

The exact values of the sine, cosine and tangent at angle 67°30′ are are 2+22 , 2−22 and 2 , respectively.

Explanation of Solution

Given information:

The given angle is 67°30′ .

Calculation:

The given angle is 67°30′ .

Convert the angle into degree.

  67°30′=67.5°

Let the given angle is 67.5° and it is half of angle 135° . Since, angle 67.5° lies in Quadrant I.So, the values of sine, cosine and tangent will be positive.

Write the expression to calculate the value of sin67.5° .

  sin67.5°=±1−cos135°2=±1−(−22)2 {cos135°=−22}

Since, the value of sine is positive in Quadrant I. So, consider the positive value and simplify it.

  sin67.5°=1−(−22)2=2+22×2=2+22

Write the expression to calculate the value of cos67.5° .

  cos67.5°=±1+cos135°2=±1+(−22)2 {cos135°=−22}

Since, the value of cosine is positive in Quadrant I. So, consider the positive value and simplify it.

  cos67.5°=1+(−22)2=2−22×2=2−22

Similarly, write the expression to calculate the value of tan67.5° .

  tan67.5°=±(1−cos135°sin135°)=±(1−(−22)(12)) {cos135°=−22sin135°=12}

Since, the value of tangent is positive in Quadrant I. So, consider the positive value and simplify it.

  tan67.5°=1−(−22)(12)=(2+22)(12)=2

Therefore, the exact values of the sine, cosine and tangent at angle 67°30′ are 2+22 ,

  2−22 and 2 , respectively.

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Chapter 5.5, Problem 51E
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Chapter 5.5, Problem 53E

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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How to apply the law of sines to find the remaining parts of a triangle; Author: Brian McLogan;https://www.youtube.com/watch?v=NdRF18HWkmE;License: Standard YouTube License, CC-BY