Question

how to find positive root of sin^-1(x) + cos^-1(4x)= pi/6

Answer 1

hmm i wonder
cant think of anything other than maybe taking the sine of both sides? maybe?

Answer 2

satellite73 u meant that by multiplying sine to both side....??

Answer 3

i think that is what you are going to have to do
\[\sin(\arcsin(x)+\arccos(4x))=\frac{1}{2}\] as a first step

Answer 4

the expression on the left can be written without trig functions via
\[\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\]

Answer 5

so arcsine(x) we assume as ALPHA n arccos(4X) as BETA is'nt

Answer 6

yes

Answer 7

@mazfar24 were you able to get solution?

Answer 8

no...:1

Answer 9

i got stuck when it become like this \[\sin \alpha \cos \beta +\cos \alpha \sin \beta =1/2\]

Answer 10

don't forgot what alpha and beta are

Answer 11

\[\sin(\arcsin(x))\cos(\arccos(4x))+\cos(\arcsin(x))\sin(\arccos(4x))=\frac{1}{2}\]

Answer 12

what is sin(arcsin(x))=?

Answer 13

what does u meant...erm, i think sin(arcsin(X)) = to x because i tried to using phatogoras theorem...am i correct

Answer 14

yes that is equal to x
what is cos(arccos(4x))=?

Answer 15

i guest it is = 4x, am i correct??

Answer 16

ok now you need to evaluate cos(arcsin(x)) and also evaluate sin(arccos(4x))
do you know how to do that?

Answer 17

\[\cos(\arcsin x) = \sqrt{1-\sin^2 (\arcsin x)} = \sqrt{1-x^2}\]

Answer 18

i think my answer is same like dumbcow...\[\sqrt{1-x ^{2}}\]

Answer 19

this is from the pythagorean identity:
sin^2 +cos^2 = 1

so similarly
\[\sin(\arccos 4x) = \sqrt{1-\cos^2 (\arccos 4x)} = \sqrt{1-16x^2}\]

Answer 20

am i correct

Answer 21

yes

Answer 22

but y ur answer is \[\sqrt{1-16x^{2}}\]

Answer 23

because you are squaring the "4x"

Answer 24

okey, that answer is for sin(arccos4x) is'nt...so then what we must do

Answer 25

yes it is, my post clearly states that

ok so now we have:
\[4x^2 + \sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2}\]
isolate sqrt term
\[\sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2} - 4x^2\]
square both sides to get rid of sqrt
\[(1-x^2)(1-16x^2) = ( \frac{1}{2} - 4x^2)^2\]
solve like you would a quadratic equation

Answer 26

am i will get this answer \[4+\sqrt{67}/34\] n [4-\sqrt{67}/34\]

Answer 27

hmm thats not what i get

what do get after distributing and combining like terms?

Answer 28

i got \[17x^{2}-4x-3/4=0\]

Answer 29

am i correct??

Answer 30

ahh not quite, that should be "4x^2" not just 4x

Answer 31

y...can u show how u get that..please...

Answer 32

\[\rightarrow 1 -x^2 -16x^2 +16x^4 = \frac{1}{4} -4x^2 +16x^4\]
\[1-17x^2 = \frac{1}{4} -4x^2\]
\[13x^2 = \frac{3}{4}\]