/*! 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 34 Use the half-angle formulas to d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the half-angle formulas to determine the exact values of the sine, cosine, and tangent of the angle. $$67^{\circ} 30^{\prime}$$

Short Answer

Expert verified
The exact values of sine, cosine, and tangent for the angle are given by \( \sin(67.5^{\circ}) = \frac{ \sqrt{2 + \sqrt{2}} }{2}\), \( \cos(67.5^{\circ}) = \frac{ \sqrt{2 - \sqrt{2}} }{2}\), and \( \tan(67.5^{\circ}) = 1 + \sqrt{2}\), respectively.

Step by step solution

01

Convert the given angle to degrees.

Here, the given angle is \(67^{\circ} 30^{\prime}\). Convert it to degrees: \(67^{\circ} 30^{\prime} = 67.5^{\circ}\).
02

Double the angle.

Double the angle so that it aligns with a well-known trigonometric value. This gives \(2 \times 67.5^{\circ} = 135^{\circ}\). Using the unit circle, we know that \(\cos(135^{\circ}) = -\frac{1}{\sqrt{2}}\).
03

Apply the half-angle formulas.

Plug these values into the half-angle formulas to find the sine, cosine, and tangent values: \( \sin(67.5^{\circ}) = \sqrt{\frac{1 - (-\frac{1}{\sqrt{2}})}{2}}\), \( \cos(67.5^{\circ}) = \sqrt{\frac{1 + (-\frac{1}{\sqrt{2}})}{2}}\), \( \tan(67.5^{\circ}) = \frac{\sin(67.5^{\circ})}{\cos(67.5^{\circ})}\). Simplify each quantity to obtain the final values.

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.