Solving an inverse trigonometric equation

[@joyli9893]

arcsin x * arccos x = pi^2 / 18

[@renyxadarox]

Suggestion: Can you evaluate cos(44.7°) ?

[@leif1075]

I did it basically the same way you did, but consider the sine sum.anglenformula sine of pi/2 plus theta equals sine pi/2costheta plus sine thetacospi/2 which simplifies to sine of pi/2 plus theta equals cosine theta and if both 0f these equal x..then inverse sine of x equals pi/2 PLUS theta..which means inverse sine of x equals pi/2 plus inverse cosine x because that3what theta by itself is..so you get that wheb you solve for pi/2 that inverse sine x MINUS inverse cosine x equals pi/2..isnt that curious??

[@SyberMath]

Great ideas! You want to be careful with inverse sin and inverse cos, though. Since y=sinx and y=cosx are not one-to-one, they don't have inverses unless you restrict their domains.

[@leif1075]

@@SyberMath I'm not sure i know what you mean..but it is still true that sine of pi/2 plus theta equals cosine theta because of the law of sum of angles ofnsines remember?

[@leif1075]

@@SyberMath but inverse sine and cosine are always defined it's not like tan x that has an asymotote ..after all even if the answer is zero like cosine of pi/2 equals sero so inverse cosine of zero is pi/2 ..or you mean since the aues can o ly be between -1 and 1 of the inverse? But that still doesnt change the fsct that the relation is true for any angles according tonthe sum of sine angles formula

[@SyberMath]

@@leif1075 This is what I mean: On the interval [0,pi] there are two angles whose sin is equal to the same thing. For example: sin(pi/6)=sin(5pi/6)=1/2. So sin inverse of 1/2 is not uniquely determined if you don't restrict the domain.

[@leif1075]

@@SyberMath ok so what i said is also correct then right..sine pi/2 plus theta equals cosine theta..thanks

[@MikeB3542]

That pi-squared over 18 feels very Basel-y!

[@lebker1342]

Guys wtf is this? |1/sin|>1 |1/cos|>1 (Pi^2)/18

[@SatyanarayanaMudunuri]

If arcsin x = A and arccos x = B then we have AB = pi^2/18 and A+B = pi/2. From here it is simple to solve. Your solution is far from elegant

[@SyberMath]

Thanks

[@Kdd160]

😀 cooool

[@SyberMath]

Thank you! 😀

[@SyberMath]

Problem fixed. Re-recorded: kzbin.info/www/bejne/iorUno2IpJWaecU

[@Kdd160]

@@SyberMath Thanks!

[@SyberMath]

@@Kdd160 You're welcome! Have a good one!

[@rex_yourbud]

My way : Notice you can break pi²/18 = ( pi/a ) ( pi/b ) where ab = 18 Now 6x3 = 18 you get rhs as pi/3 times pi/6, which are complementary angles and super nice. First equate arc(sin) to pi/6 and then to pi/3 and arc(cos) will take care of itself since angles are complimentary. Boom the two solution : 1/sqrt(2) and sqrt(3)/2

[@johncena-fy3mo]

I. Also got a very challenging problem can you please help me out?

[@SyberMath]

I'm not sure if I will be able to help but feel free to send a tweet and mention me @SyberMath. If I cannot solve it, I will pose it on twitter and someone will probably figure it out!

[@johncena-fy3mo]

@@SyberMath I could send u on instagram

[@SyberMath]

Sure.

[@SyberMath]

Problem fixed. Re-recorded: kzbin.info/www/bejne/iorUno2IpJWaecU