Question

Sin^-1 (1/3) + Sin^-1 [2/3 (1 - 1/ root8]

Solve it..

IIT FOUNDATION - Mathematics - Class 9th and 10th.

Answer 1

Hello!

Solve it:

$Sin^{^{-1}}\:\left(\dfrac{1}{3}\right)\:+\:Sin^{^{-1}}\:\left[\frac{2}{3}\:\left(1\:-\:\frac{1}{\sqrt{8}}\right)\right]$

Solution:

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{2}{3}\:\left(1\:-\:\dfrac{1}{\sqrt{8}}\right)\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{2}{3}\:\left(1\:-\:\dfrac{1}{\sqrt{2^3}}\right)\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{2}{3}\:\left(1\:-\:\dfrac{1}{\sqrt{2^2*2}}\right)\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{2}{3}\:\left(1\:-\:\dfrac{1}{2\sqrt{2}}\right)\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{2}{3}\:\left(\dfrac{2\sqrt{2}-1}{2\sqrt{2}}\right)\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{\diagup\!\!\!\!2*\left(\dfrac{2\sqrt{2}-1}{\diagup\!\!\!\!2\sqrt{2}}\right)}{3}\:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\dfrac{\left(\dfrac{2\sqrt{2}-1}{\sqrt{2}}\right)}{3}\:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{2\sqrt{2}-1}{\sqrt{2}}*\dfrac{1}{3} \right)}\:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{2\sqrt{2}-1}{3\sqrt{2}} \right)}\:\right]$

multiply the second part by $\dfrac{\sqrt{2} }{\sqrt{2} }$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{2\sqrt{2}-1}{3\sqrt{2}} \right)}*\dfrac{\sqrt{2} }{\sqrt{2} } \:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{\sqrt{2}*2\sqrt{2}-1*\sqrt{2} }{\sqrt{2}*3\sqrt{2}} \right)} \:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{2\left(\sqrt{2}\right)^2-\sqrt{2} }{\left(\sqrt{2}\right)^2*3 \right)} \:\right]$

$arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left[\left(\dfrac{2*2-\sqrt{2} }{2*3 \right)} \:\right]$

$\boxed{\boxed{arcsin\:\left(\dfrac{1}{3}\right)\:+\:arcsin\:\left(\dfrac{4-\sqrt{2} }{6 \right)}}}\Longleftarrow(answer)$

or

$\boxed{\boxed{decimal = 0.78540...}}\Longleftarrow(answer)$

or

$\boxed{\boxed{degree = 45\º}}\Longleftarrow(answer)$

__________________________

I Hope this helps, greetings ... Dexteright02! =)

Answer 2

in inverse trigonometry, don't memorise so much formula . just learn trigonometry identities , you can solve it .

Let's try to solve it,
$sin^{-1}\frac{1}{3}+sin^{-1}[\frac{2}{3}(1-\frac{1}{\sqrt{8}})]$

Let sin^-1(1/3) = A => sinA = 1/3
cosA = √8/3
sin^-1[2/3(1 - 1/√8)] = B => sinB = 2/3[1 - 1/√8]
sinB = 2(√8 - 1)/3√8 = 2(2√2-1)/6√2
sinB = (2√2 - 1)/3√2 = (4 - √2)/6
so, cosB = √{6² - (4-√2)²}/6
= √{36 -(16 + 2 - 8√2)}/6
= √{18 + 8√2}/6
= √{4² + √2² + 2.4.√2}/6
= √(4 + √2)²/6
= (4 + √2)/6
= (2√2 + 1)/3√2
= 2(√8 + 1)/3√8
= 2/3(1 + 1/√8)
hence, cosB = 2/3(1 + 1/√8)

sin(A + B) = sinA.cosB + cosA.sinB
= 1/3 × 2/3(1 + 1/√8) + √8/3 × 2/3(1 - 1/√8)
= 2/9 + 2/9√8 + 2√8/9 - 2/9
= 2/9[1/√8 + √8]
now, A + B = sin^-1 [2/9(1/√8 + √8)]
e.g., sin^-1(1/3) + sin^-1[2/3(1-1/√8)] = sin^-1[2/9(1/√8+√8)]