$cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)$ true for all $x$?












0












$begingroup$



$cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)$ true for all $x$ ?




My Attempt



Let $a=cos^{-1}x$, $b=cos^{-1}yimplies$$cos a=x$, $cos b=y$ and $a,bin[0,pi]implies a+bin[0,2pi]$
$$
cos(a+b)=cos acos b-sin asin b=xy-sqrt{1-x^2}sqrt{1-y^2}\=cosbigg[cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)bigg]\
a+b=color{red}{cos^{-1}x+cos^{-1}y=2npipmcos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)}
$$
Case 1: $a+bin[0,pi)$
$$
cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$



Case 2: $a+bin[pi,2pi]$
$$
cos^{-1}x+cos^{-1}y=2pi-cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$
Case 1-: $a+bin[0,pi)$



Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.






trigonometry inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3027787/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 6:46










  • $begingroup$
    @labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
    $endgroup$
    – ss1729
    Dec 6 '18 at 6:47












  • $begingroup$
    For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 6:53










  • $begingroup$
    @Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:12












  • $begingroup$
    @labbhattacharjee could you please comment on my attempt ?
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:16
















0












$begingroup$



$cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)$ true for all $x$ ?




My Attempt



Let $a=cos^{-1}x$, $b=cos^{-1}yimplies$$cos a=x$, $cos b=y$ and $a,bin[0,pi]implies a+bin[0,2pi]$
$$
cos(a+b)=cos acos b-sin asin b=xy-sqrt{1-x^2}sqrt{1-y^2}\=cosbigg[cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)bigg]\
a+b=color{red}{cos^{-1}x+cos^{-1}y=2npipmcos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)}
$$
Case 1: $a+bin[0,pi)$
$$
cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$



Case 2: $a+bin[pi,2pi]$
$$
cos^{-1}x+cos^{-1}y=2pi-cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$
Case 1-: $a+bin[0,pi)$



Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.






trigonometry inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3027787/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 6:46










  • $begingroup$
    @labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
    $endgroup$
    – ss1729
    Dec 6 '18 at 6:47












  • $begingroup$
    For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 6:53










  • $begingroup$
    @Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:12












  • $begingroup$
    @labbhattacharjee could you please comment on my attempt ?
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:16














0












0








0





$begingroup$



$cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)$ true for all $x$ ?




My Attempt



Let $a=cos^{-1}x$, $b=cos^{-1}yimplies$$cos a=x$, $cos b=y$ and $a,bin[0,pi]implies a+bin[0,2pi]$
$$
cos(a+b)=cos acos b-sin asin b=xy-sqrt{1-x^2}sqrt{1-y^2}\=cosbigg[cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)bigg]\
a+b=color{red}{cos^{-1}x+cos^{-1}y=2npipmcos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)}
$$
Case 1: $a+bin[0,pi)$
$$
cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$



Case 2: $a+bin[pi,2pi]$
$$
cos^{-1}x+cos^{-1}y=2pi-cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$
Case 1-: $a+bin[0,pi)$



Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.






trigonometry inverse-function







share|cite|improve this question











$endgroup$





$cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)$ true for all $x$ ?




My Attempt



Let $a=cos^{-1}x$, $b=cos^{-1}yimplies$$cos a=x$, $cos b=y$ and $a,bin[0,pi]implies a+bin[0,2pi]$
$$
cos(a+b)=cos acos b-sin asin b=xy-sqrt{1-x^2}sqrt{1-y^2}\=cosbigg[cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)bigg]\
a+b=color{red}{cos^{-1}x+cos^{-1}y=2npipmcos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)}
$$
Case 1: $a+bin[0,pi)$
$$
cos^{-1}x+cos^{-1}y=cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$



Case 2: $a+bin[pi,2pi]$
$$
cos^{-1}x+cos^{-1}y=2pi-cos^{-1}Big(xy-sqrt{1-x^2}sqrt{1-y^2}Big)
$$
Case 1-: $a+bin[0,pi)$



Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.






trigonometry inverse-function




trigonometry inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 7:13







ss1729

















asked Dec 6 '18 at 6:45









ss1729ss1729

2,00111024




2,00111024












  • $begingroup$
    See math.stackexchange.com/questions/3027787/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 6:46










  • $begingroup$
    @labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
    $endgroup$
    – ss1729
    Dec 6 '18 at 6:47












  • $begingroup$
    For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 6:53










  • $begingroup$
    @Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:12












  • $begingroup$
    @labbhattacharjee could you please comment on my attempt ?
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:16


















  • $begingroup$
    See math.stackexchange.com/questions/3027787/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 6:46










  • $begingroup$
    @labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
    $endgroup$
    – ss1729
    Dec 6 '18 at 6:47












  • $begingroup$
    For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
    $endgroup$
    – Andrei
    Dec 6 '18 at 6:53










  • $begingroup$
    @Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:12












  • $begingroup$
    @labbhattacharjee could you please comment on my attempt ?
    $endgroup$
    – ss1729
    Dec 6 '18 at 7:16
















$begingroup$
See math.stackexchange.com/questions/3027787/…
$endgroup$
– lab bhattacharjee
Dec 6 '18 at 6:46




$begingroup$
See math.stackexchange.com/questions/3027787/…
$endgroup$
– lab bhattacharjee
Dec 6 '18 at 6:46












$begingroup$
@labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
$endgroup$
– ss1729
Dec 6 '18 at 6:47






$begingroup$
@labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $cos^{-1}x+cos^{-1}y$ on internet.
$endgroup$
– ss1729
Dec 6 '18 at 6:47














$begingroup$
For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
$endgroup$
– Andrei
Dec 6 '18 at 6:53




$begingroup$
For $x=y=-1$, the LHS is $2pi$, the RHS is $0$.
$endgroup$
– Andrei
Dec 6 '18 at 6:53












$begingroup$
@Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
$endgroup$
– ss1729
Dec 6 '18 at 7:12






$begingroup$
@Andrei For $x=y=-1implies a=b=pi$. RHS=$2pi-cos^{-1}1=2pi=LHS$, it is case 2. srry abt that, just edited the domain.
$endgroup$
– ss1729
Dec 6 '18 at 7:12














$begingroup$
@labbhattacharjee could you please comment on my attempt ?
$endgroup$
– ss1729
Dec 6 '18 at 7:16




$begingroup$
@labbhattacharjee could you please comment on my attempt ?
$endgroup$
– ss1729
Dec 6 '18 at 7:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

Using Principal values and Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$,



$cos^{-1}x+cos^{-1}yle2pi$



$$cos^{-1}x+cos^{-1}y=cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}ylepi$



$iffsin^{-1}x+sin^{-1}yge0iffsin^{-1}xge-sin^{-1}y=sin^{-1}(-y)$



$iff xge -yiff x+yge0$



$$cos^{-1}x+cos^{-1}y=2pi-cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}y>pi$



$iffsin^{-1}x+sin^{-1}y<0iff x+y<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
    $endgroup$
    – ss1729
    Dec 6 '18 at 17:48








  • 1




    $begingroup$
    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 18:24











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Using Principal values and Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$,



$cos^{-1}x+cos^{-1}yle2pi$



$$cos^{-1}x+cos^{-1}y=cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}ylepi$



$iffsin^{-1}x+sin^{-1}yge0iffsin^{-1}xge-sin^{-1}y=sin^{-1}(-y)$



$iff xge -yiff x+yge0$



$$cos^{-1}x+cos^{-1}y=2pi-cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}y>pi$



$iffsin^{-1}x+sin^{-1}y<0iff x+y<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
    $endgroup$
    – ss1729
    Dec 6 '18 at 17:48








  • 1




    $begingroup$
    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 18:24
















1












$begingroup$

Using Principal values and Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$,



$cos^{-1}x+cos^{-1}yle2pi$



$$cos^{-1}x+cos^{-1}y=cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}ylepi$



$iffsin^{-1}x+sin^{-1}yge0iffsin^{-1}xge-sin^{-1}y=sin^{-1}(-y)$



$iff xge -yiff x+yge0$



$$cos^{-1}x+cos^{-1}y=2pi-cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}y>pi$



$iffsin^{-1}x+sin^{-1}y<0iff x+y<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
    $endgroup$
    – ss1729
    Dec 6 '18 at 17:48








  • 1




    $begingroup$
    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 18:24














1












1








1





$begingroup$

Using Principal values and Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$,



$cos^{-1}x+cos^{-1}yle2pi$



$$cos^{-1}x+cos^{-1}y=cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}ylepi$



$iffsin^{-1}x+sin^{-1}yge0iffsin^{-1}xge-sin^{-1}y=sin^{-1}(-y)$



$iff xge -yiff x+yge0$



$$cos^{-1}x+cos^{-1}y=2pi-cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}y>pi$



$iffsin^{-1}x+sin^{-1}y<0iff x+y<0$






share|cite|improve this answer









$endgroup$



Using Principal values and Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$,



$cos^{-1}x+cos^{-1}yle2pi$



$$cos^{-1}x+cos^{-1}y=cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}ylepi$



$iffsin^{-1}x+sin^{-1}yge0iffsin^{-1}xge-sin^{-1}y=sin^{-1}(-y)$



$iff xge -yiff x+yge0$



$$cos^{-1}x+cos^{-1}y=2pi-cos^{-1}(xy-sqrt{(1-x^2)(1-y^2)})$$



will hold true if $cos^{-1}x+cos^{-1}y>pi$



$iffsin^{-1}x+sin^{-1}y<0iff x+y<0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 7:33









lab bhattacharjeelab bhattacharjee

226k15157275




226k15157275












  • $begingroup$
    Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
    $endgroup$
    – ss1729
    Dec 6 '18 at 17:48








  • 1




    $begingroup$
    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 18:24


















  • $begingroup$
    Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
    $endgroup$
    – ss1729
    Dec 6 '18 at 17:48








  • 1




    $begingroup$
    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 18:24
















$begingroup$
Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
$endgroup$
– ss1729
Dec 6 '18 at 17:48






$begingroup$
Thanks. I just have one more query. Could u please confirm that, $cos^{-1}x-cos^{-1}y=begin{cases}cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-yleq 0\ -2pi+cos^{-1}bigg(xy+sqrt{1-x^2}sqrt{1-y^2}bigg),;x-y> 0end{cases}$
$endgroup$
– ss1729
Dec 6 '18 at 17:48






1




1




$begingroup$
@ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
$endgroup$
– lab bhattacharjee
Dec 6 '18 at 18:24




$begingroup$
@ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use math.stackexchange.com/questions/1224415/…
$endgroup$
– lab bhattacharjee
Dec 6 '18 at 18:24


















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How to send String Array data to Server using php in android

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0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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Is anime1.com a legal site for watching anime?

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up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
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