Question

Prove that :cos inverse (x) + sin inverse (x/2) = pi/6

Answer 1

you mean you have to solve for 'x' ? right ?
(and not "prove" that its true for all values of x)

Answer 2

because its clearly not true for x=0

Answer 3

yeah, sry abt that....it's to find x....

Answer 4

good, by inspection, i could find one value of x, but lemme think on how to do it algebraically...

Answer 5

oh, well okay....but how by inspection?

Answer 6

is that a multiple choice question ?

Answer 7

nope....

Answer 8

well, because i knew that cos inverse of 1 =0
so i found out that sin inverse of 1/2 is indeed pi/6
which gave me x=1 as one solution

Answer 9

oh, yes....actually :D
you're right sir, didn't spark in...

Answer 10

just checked with calculator, there are no other solutions.
do you require it to be solved algebraically ?

i've tried substituting, x= 2 sin theta, but as of now it led me to nowhere....

Answer 11

well it's a long answer type question....

Answer 12

i wish we had a standard formula for sin inverse of sin 2theta

trying triangular method now...

Answer 13

lol, sounds good....

Answer 14

To prove the identity...we can replace the left-hand side by their corresponding angle such as
\[\theta = \cos^{-1} x\]
\[\alpha=\sin^{-1} \frac{ x }{ 2 }\]
therefore
\[\theta+\alpha=\frac{ \pi }{ 6 }\]

Answer 15

Applying Sine at both sides of the equations...
\[\sin(\theta+\alpha)=\sin \frac{ \pi }{ 6 }\]

Answer 16

applying Addition formula for Sine at left-hand side and evaluate sin (pi/6) = 1/2...
\[\sin \theta \cos \alpha + \cos \theta \sin \alpha = \frac {1}{2}\]

Answer 17

okay, and then....

Answer 18

what is sin theta?

\[\sin \theta = \sqrt{1-x^2}\]

Answer 19

what is cos alpha?

\[\cos \alpha = \frac{ \sqrt{1-x^2} }{ 2 }\]

Answer 20

substitute it to the Sine Addition of theta and alpha... to review
cos theta = x
sin theta = sqrt(1-x^2)
cos alpha = sqrt(1-x^2)/2
sin alpha = x/2

Answer 21

\[\sqrt{1-x^2}\times \frac{ \sqrt{1-x^2} }{ 2 }+x \times \frac{ x }{ 2 }=\frac{ 1 }{ 2 }\]

Answer 22

if you will continue, it will lead you to the equality of both sides...

Answer 23

yup i know.. i did not notice it. thanks for the correction, a topo error from mine... :)

Answer 24

that proves the identity... :)

Answer 25

umm, actually.... okay...
thanx...

Answer 26

got it thanx....

Answer 27

no problem... just always looking around... :)

Answer 28

\(\sqrt{1-x^2}\times \sqrt{1-(x/2)^2}+x \times \frac{ x }{ 2 }=\frac{ 1 }{ 2 }\)
does not prove the identity, infact its not an identity at all as i already mentioned...

Answer 29

ok stand me corrected... not an identity... simply an inverse trigonometric relationship... :)

Answer 30

thanks @hartnn for the comments... :)

Answer 31

i don't see any easy way to solve for x

Answer 32

Umm, this isnt fitting in @Orion1213 ....

Answer 33

the cos alpha part.....plz check...

Answer 34

ok... we do not intend to solve the x value here... what we are trying to show is that both sides of the equation are equal... that is

cos inverse (x) + sin inverse (x/2) = pi/6

Answer 35

i use alpha = sin inverse (x/2)
so sin alpha = x/2
our cos alpha = sqrt[1-(x/2)^2] = sqrt[1-(x^2)/4)] = sqrt(1-x^2)/2

Answer 36

back to...
\[\sin(\theta+\alpha) = \sin \frac{\pi}{6} = \frac{1}{2}\]
\[\sin \theta \cos \alpha + \sin \alpha \cos \theta = \frac{1}{2}\]
\[\sqrt{1-x^2}\times \frac{\sqrt{1-x^2}}{2}+\frac{x}{2}\times x = \frac{1}{2}\]
we can cancel (1/2) on both sides...
\[(\sqrt{1-x^2})^2+x^2=1\]
\[(1-x^2)+x^2=1\]
or simply...
\[1=1\]
... proven to be equal...

Answer 37

you might find it slightly easier to solve this by first rearranging as follows:\[\begin{align}
\cos^{-1}(x)+\sin^{-1}(x/2)&=\pi/6\\
\cos^{-1}(x)&=\pi/6-\sin^{-1}(x/2)\\
\cos(\cos^{-1}(x))&=\cos(\pi/6-\sin^{-1}(x/2))\\
x&=\cos(\pi/6)\cos(\sin^{-1}(x/2))+\sin(pi/6)\sin(\sin^{-1}(x/2))\\
x&=(\sqrt{3}/2)\sqrt{1-x^2/4}+(1/2)(x/2)\\
&=(\sqrt{3}/4)\sqrt{4-x^2}+x/4\\
4x&=\sqrt{12-3x^2}+x\\
3x&=\sqrt{12-3x^2}\\
9x^2&=12-3x^2\\
12x^2&=12\\
x&=1
\end{align}\]

Answer 38

ok got the method @asnaseer ... thanks i learned a lot from this problem... thanks again to @hartnn... :)

Answer 39

yw :)

Answer 40

thanks @anubhav.m ... :)

Answer 41

sure @Orion1213 , thanx for the help evryone....especially @asnaseer ..... :)