Chapter 5: Problem 41
Show that \(\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}\) [Hint: \(15=45-30]\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 5: Problem 41
Show that \(\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}\) [Hint: \(15=45-30]\)
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Find the smallest positive number \(x\) such that $$ \sin ^{2} x-3 \sin x+1=0 $$
Evaluate \(\sin \left(\sin ^{-1}\left(\frac{1}{e}-\frac{1}{\pi}\right)\right)\)
Show that $$ \cos \left(\tan ^{-1} t\right)=\frac{1}{\sqrt{1+t^{2}}} $$ for every number \(t\).
Show that if \(t<0,\) then $$ \tan ^{-1} \frac{1}{t}=-\frac{\pi}{2}-\tan ^{-1} t. $$
Show that \(\cos (3 \theta)=4 \cos ^{3} \theta-3 \cos \theta\) for all \(\theta\). \([\operatorname{Hint}: \cos (3 \theta)=\cos (2 \theta+\theta) .]\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.