If-sin-1-sin-sin-sin-1-sin-sin-pi-2-find-the-value-of-sin-2-sin-2-1-2-

Question Number 137530 by ZiYangLee last updated on 03/Apr/21

If \sin^{- 1} (\sin α + \sin β) + \sin^{- 1} (\sin α - \sin β) = \frac{π}{2}

find the value of \sin^{2} α + \sin^{2} β . [\frac{1}{2}]

Answered by mr W last updated on 04/Apr/21

\sin α + \sin β = u

\sin α - \sin β = v

\sin^{2} α + \sin^{2} β = \frac{u^{2} + v^{2}}{2}

\sin^{- 1} u + \sin^{- 1} v = \frac{π}{2}

\sin^{- 1} u = \frac{π}{2} - \sin^{- 1} v

u = \sin (\frac{π}{2} - \sin^{- 1} v) = \cos (\sin^{- 1} v)

= \sqrt{1 - \sin^{2} (\sin^{- 1} v)} = \sqrt{1 - v^{2}}

\Rightarrow u^{2} = 1 - v^{2}

\Rightarrow u^{2} + v^{2} = 1

\Rightarrow \sin^{2} α + \sin^{2} β = \frac{u^{2} + v^{2}}{2} = \frac{1}{2}

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

prof i want to learn something from

you can you please prove the

\sin^{2} A + \sin^{2} B = 1 for me ?

i only know of \sin^{2} A + \cos^{2} A = 1

sorry for worrying you

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

wow!

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

very nice solution but am finding it

difficult to understand where the \cos

and the square root is introduce

please help me

Commented by mr W last updated on 04/Apr/21

s a y

A = \sin^{- 1} u \Rightarrow \sin A = u

B = \sin^{- 1} v \Rightarrow \sin B = v

A + B = \frac{π}{2}

A = \frac{π}{2} - B

\sin A = \sin (\frac{π}{2} - B) = \cos B = \sqrt{1 - \sin^{2} B}

\sin^{2} A = 1 - \sin^{2} B

\sin^{2} A + \sin^{2} B = 1

u^{2} + v^{2} = 1

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

now is very clear God bless you alot

for the kindness profW

Commented by mr W last updated on 04/Apr/21

\sin^{2} A + \cos^{2} A = 1 ✓

s i n c e A = \frac{π}{2} - B, w h i c h m e a n s

\cos A = \sin B, s o w e g e t

\sin^{2} A + \sin^{2} B = 1 ✓

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

God bless u profW

Commented by ZiYangLee last updated on 10/Apr/21

ProfW, how can

\sqrt{1 - \sin^{2} (\sin^{- 1} b)} = \sqrt{1 - v^{2}}

Commented by mr W last updated on 16/Apr/21

y o u m e a n

\sqrt{1 - \sin^{2} (\sin^{- 1} v)} = \sqrt{1 - v^{2}} ?

s a y \sin^{- 1} v = V, t h a t m e a n s

\sin V = v

i . e . \sin (\sin^{- 1} v) = v

s o \sin^{2} (\sin^{- 1} v) = v^{2}

\sqrt{1 - \sin^{2} (\sin^{- 1} v)} = \sqrt{1 - v^{2}}

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