Question

If sin( sin inverse 1/5 + cos inverse x) = 1

Find the value of x

Answer 1

you may use sin(A+B) formula

Answer 2

That is sin(A + B) = sinAcosB + SinBcosA?

Answer 3

@rational Uhm if SinA is sin^-1 1/5 what is SinB?

Answer 4

i'd focus on the fact that sin(....)=1, specifically the fact that it equals 1

Answer 5

So we get it into a sin^2 x + cos^2 x form?

Answer 6

\(\sin(\sin^{-1}x) = x\)

Answer 7

look at it this way: \[if \sin^{-1} \frac{ 1 }{ 5 }\rightarrow \sin y=\frac{ 1 }{ 5 }, and \cos^{-1} x \rightarrow \cos z=x\]

Answer 8

Oh okay so if

sin y = 1/5

cos z = x

I write it as Sin(sin y 1/5 . cos z + sin z 1/5 . cos x) = 1?

Answer 9

oh nice trick

Answer 10

use the sin(A+B)

Answer 11

sec writing it in my book

Answer 12

Alternatively if you can recall some inverse identities, start like this\[\sin^{-1}(1/5) + \cos^{-1}(x) = \dfrac{\pi}{2}\]

Answer 13

(assuming principle angles.. )

Answer 14

Notice that x=1/5 satisfies above equation if you remember the identity
\[\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}\]

Answer 15

I got this so far..

sin(sin^-1 1/5).cos(cos^-1 x) + sin(cos^-1 x).cos(sin^-1 1/5) = 1

And that = 1/5 . x + sin(cos^-1 x).cos(sin^-1 1/5)

Answer 16

Yeah I'm stuck. How would you use the pi/2 when cos inverse and sin inverse are locked in the round brackets?

Answer 17

fine so far, next eliminate trig completely..
cos(sin^-1 1/5) = sqrt(24)/5

Answer 18

sin(cos^-1 x) = sqrt(1-x^2)

Answer 19

How'd you get sqrt(24)/5?

Answer 20

\(\sin^{-1}(1/5) = t \implies \sin t = 1/5 \implies \cos t = \sqrt{1-(1/5)^2} = \sqrt{24}/5\)

Answer 21

Oh.. oooh o.

Okay I get it!

hang on, lemme try completing the rest of the problem

Answer 22

Okay... rest is just algebra

Answer 23

$$
\sin\left ( \sin^{-1}{1/5 }+ \cos ^{-1} x\right ) = 1\\
\sin^{-1}\sin(\left ( \sin^{-1}{1/5 }+ \cos ^{}-1 x\right ))=\sin^{-1}1={\pi\over2}\\
\cos ^{-1} x={\pi\over2}-\sin^{-1}{1\over 5 }\\
x=\cos\left({\pi\over2}-\sin^{-1}{1\over 5 }\right)\\
$$
Does this help?

Answer 24

Got it.. finally its 1/5

the inverse cancels sin and cos

x/5 + root1 + x^2.root24/5 = 1

multiplying both sides by 5 and subtracting x

we get root 24/5.root 1 - x^2 = 5 - x

squaring both sides

2.6.1-x^2 = 25 - x^2 - 25 - 10x

or 24 - 24x^2 = 25 + x^2 - 10x

or 25x^2 + 1 + 10x = 0

(5x -1)^2 = 0

5x - 1 = 0

x = 1/5

Answer 25

Thanks everyone :) @ybarrap @rational @dan815

Answer 26

Wow! you managed to sovle that equation congrats!!
look at ybarrap's reply, it is the shortest and most straightforward solution