Question 1102370
.
If cos(t)=-5/6 where pi < t < 3pi/2 , find the values of the following trigonometric functions.
 
cos(2t) =  
sin(2t)=  
cos(t/2)=  
sin(t/2)= 
~~~~~~~~~~~~~~~~


<pre>
You are given that  cos(t) = {{{-5/6}}}  and the angle  "t"  is in  QIII.


Then   sin(t) =  {{{-sqrt(1-cos^2(t))}}} = {{{-sqrt(1-(-5/6)^2)}}} = {{{-sqrt((36-25)/36)}}} = {{{-sqrt(11/36)}}} = {{{-sqrt(11)/6}}}.


        Notice that I put the sign "-" (minus) at sqrt, since the function sine is NEGATIVE in QIII.


The next steps are straightforward.  

a)  cos(2t) = cos^2(t) - sin^2(t) = {{{(-5/6)^2 - (-sqrt(11)/6)^2}}} = {{{25/36 - 11/36}}} = {{{14/36}}} = {{{7/18}}}.


b)  sin(2t) = 2*sin(t)*cos(t) = {{{2*(-sqrt(11)/6)*(-5/6)}}} = {{{(5*sqrt(11))/18}}}.


c)   cos(t/2) = {{{-sqrt((1+cos(t))/2)}}} = {{{-sqrt((1+(-5/6))/2)}}} = {{{-sqrt((6-5)/(6*2))}}} = {{{-sqrt(1/12)}}} = {{{-sqrt(12)/12}}} = {{{(-4*sqrt(3))/12)}}} = {{{-sqrt(3)/3}}}.


         Again, the sign is "-" at sqrt, since cosine is NEGATIVE function in QII, where the angle t/2 is.


d)  sin(t/2) = {{{sqrt((1-cos(t))/2)}}} = {{{sqrt((1-(-5/6))/2)}}} = {{{sqrt((6+5)/(6*2))}}} = {{{sqrt(11/12)}}} = {{{sqrt(132)/12}}} = {{{(2*sqrt(33))/12)}}} = {{{sqrt(33)/6}}}.


         The sign is "+" at sqrt, since the sine function is POSITIVE in QII, where the angle t/2 is.
</pre>


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There are two useful groups of lessons in this site, relevant to this problem.


First group is the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Compendium-of-Trigonometry-Formulas.lesson>FORMULAS FOR TRIGONOMETRIC FUNCTIONS</A>

and all associated lessons (links) with it.


The second group is these lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Calculating-trigonometric-functions-of-angles.lesson>Calculating trigonometric functions of angles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Selected-problems-from-the-archive-on-calculating-trig-functions-of-angles.lesson>Advanced problems on calculating trigonometric functions of angles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Evaluating-trigonometric-expressions.lesson>Evaluating trigonometric expressions</A> 

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the parts of this online textbook under the topics

"<U>Trigonometry. Formulas for trigonometric functions</U>" &nbsp;&nbsp;and  &nbsp;&nbsp;"<U>Trigonometry: Solved problems</U>".



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.