Question 1052901
.
Find the exact value of the​ following, without using a calculator.
sin(sin^-1 (1/2)+tan^-1 (-5))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
0.  Let {{{alpha}}} be {{{sin^(-1)(1/2)}}} = {{{arcsin(1/2)}}}, and

    let {{{beta}}} be {{{tan(-1)(-5)}}} = {{{arctan(-5)}}}.

    Then we have {{{sin(alpha)}}} = {{{1/2}}} with {{{alpha}}} in QI and {{{tan(beta)}}} = -5 with {{{beta}}} in QIV, and we need to find {{{sin(alpha + beta)}}}.

    Surely, we will use the formula

    {{{sin(alpha + beta)}}} = {{{sin(alpha)*cos(beta) + cos(alpha)*sin(beta)}}},                 (1)

    but first we shall calculate {{{cos(beta)}}}, {{{cos(alpha)}}} and {{{sin(beta)}}}.


1.  Since {{{sin(alpha)}}} = {{{1/2}}}, we have {{{alpha}}} = {{{pi/6}}} and {{{cos(alpha)}}} = {{{sqrt(3)/2}}}.    (2)


2.  {{{sin^2(beta)}}} = {{{1/(1+1/tan^2(beta))}}} = {{{1/(1 + 1/(-5)^2)}}} = {{{25/(25+1)}}} = {{{25/26}}}.

    Hence, {{{sin(beta)}}} = {{{-sqrt(25/26)}}} = {{{-5/sqrt(26)}}}.                            (3)

    We chose the sign "-" at the square root, since the angle {{{beta}}} lies in QIV (due to the sign of tan).


3.  Since {{{sin(beta)}}} = {{{-5/sqrt(26)}}}, we have 

    {{{cos(beta)}}} = {{{sqrt(1-sin^2(beta))}}} = {{{sqrt(1-(-5/sqrt(26))^2)}}} = {{{sqrt(1-25/26)}}} = {{{sqrt((26-25)/26))}}} = {{{sqrt(1/26)}}} = {{{1/sqrt(26)}}}.    (4)


4.  Now you have everything to use the formula (1):

    {{{sin(alpha+beta)}}} = {{{(1/2)*(1/sqrt(26)) + (sqrt(3)/2)*(-5/sqrt(26))}}} = {{{(1-5*sqrt(3))/(2*sqrt(26))}}}.
</pre>

Solved.


For other similar solved problems see the lesson 

&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>

in this site.


Also, 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 lesson is the part of this online textbook under the topic "<U>Trigonometry: Solved problems</U>".