Question

pls help... inverse trigonometry question

Answer 1

ques no. 27

Answer 2

@Mehek14 @sweetburger

Answer 3

@Hero

Answer 4

The thing to notice here is that there are two copies of \(\Large \cos^{-1}(x)\) and \(\Large \sin^{-1}(y)\)

This will help us when it comes to substitution

Answer 5

but there are 3 variables... how substitution will help?

Answer 6

Focus on equation 2
\[\Large \left(\cos^{-1}(x)\right)\left(\sin^{-1}(y)\right)^2 = \frac{\pi^4}{16}\]

isolate \(\Large \left(\sin^{-1}(y)\right)^2\) to get

\[\Large \left(\sin^{-1}(y)\right)^2 = \frac{\pi^4}{16\cos^{-1}(x)}\]

with me so far?

Answer 7

yup..

Answer 8

Now move back to equation 1
\[\Large \cos^{-1}(x)+\left(\sin^{-1}(y)\right)^2=\frac{p\pi^2}{4}\]

Answer 9

Replace the \[\Large \left(\sin^{-1}(y)\right)^2\] with \[\Large \frac{\pi^4}{16\cos^{-1}(x)}\]

Answer 10

So we go from
\[\Large \cos^{-1}(x)+\left(\sin^{-1}(y)\right)^2=\frac{p\pi^2}{4}\]
to
\[\Large \cos^{-1}(x)+\frac{\pi^4}{16\cos^{-1}(x)}=\frac{p\pi^2}{4}\]

Answer 11

Agreed? or no?

Answer 12

agree..

Answer 13

ok at this point, the arccosines are a bother to me. They are more complicated than I want them to be. So what I'm going to do is let \(\Large z = \cos^{-1}(x)\)

Answer 14

ok

Answer 15

Which means
\[\Large \cos^{-1}(x)+\frac{\pi^4}{16\cos^{-1}(x)}=\frac{p\pi^2}{4}\]
becomes
\[\Large z+\frac{\pi^4}{16z}=\frac{p\pi^2}{4}\]

Answer 16

The fractions are also a pain, so we can eliminate them by multiplying every term by the LCD 16z

\[\Large z+\frac{\pi^4}{16z}=\frac{p\pi^2}{4}\]

\[\Large {\color{red}{16z*}}z+{\color{red}{16z*}}\frac{\pi^4}{16z}={\color{red}{16z*}}\frac{p\pi^2}{4}\]


\[\Large 16z^2 + \pi^4 = 4p\pi^2*z\]


agreed so far? any questions at this point?

Answer 17

ok fixed

Answer 18

i got it .. then find the roots of z

Answer 19

we don't actually need to find the actual roots. We just need to figure out which values of p yield real roots for z

Answer 20

after this point, get everything to one side
\[\Large 16z^2 + \pi^4 = 4p\pi^2*z\]
\[\Large 16z^2 + \pi^4 - 4p\pi^2*z=0\]
\[\Large 16z^2 - 4p\pi^2*z + \pi^4=0\]
Notice how the left hand side is in the form
az^2 + bz + c = 0
where...
a = 16
b = -4p*pi^2
c = pi^4

Answer 21

i think ..p should be 2 because it will then make
(4z - pi^2 )^2 equation

Answer 22

The equation az^2 + bz + c = 0 only has real roots for z if and only if D >= 0
where

D = b^2 - 4ac
this is the discriminant formula


So let's plug the values in


D = b^2 - 4ac
D = (-4p*pi^2)^2 - 4*16*pi^4
D = 16p^2*pi^4 - 64pi^4


D >= 0
16p^2*pi^4 - 64pi^4 >= 0
pi^4*(16p^2 - 64) >= 0
16p^2 - 64 >= 0
16p^2 >= 64
p^2 >= 4
p <= -2 or p >= 2



If p <= -2 or p >= 2, the discriminant D is positive

If p <= -2 or p >= 2, then the original system will have real solutions.


I agree, the only possible solution in your answer choices is B) 2.

Answer 23

thanks a lot for ur explanation

Answer 24

no problem

Answer 25

I'm not sure why, but you could guess the solution pretty easily. You can't help but notice that both equations are suggesting \(\sin^{-1}(y)\) to be \(\pi/2\). That also makes \(\cos^{-1 }(x) = \frac{\pi^2}{4}\)

Answer 26

yup.. u are right but i wanna see a positive approach too