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Question Number 196166 by pticantor last updated on 19/Aug/23

f o r a n y a, b \in [- 1, 1] c a l c u l a t e :

a r c c o s (a) + a r c s i n (b) = ?

a r c s i n (a) + a r c s i n (b) = ?

a r c c o s (a) + a r c c o s (b) = ?

Commented by pticantor last updated on 19/Aug/23

i m a k e s o m e u n f o r g o t a b l e m i s t a k e! p l e a s e

Answered by MM42 last updated on 19/Aug/23

1) {c o s}^{- 1} a = x \Rightarrow a = c o s x \Rightarrow s i n x = \sqrt{1 - a^{2}}

{s i n}^{- 1} b = y \Rightarrow b = s i n y \Rightarrow c o s y = \sqrt{1 - b^{2}}

A = x + y \Rightarrow s i n A = s i n x c o s y + s i n y c o s x

\Rightarrow s i n A = \sqrt{1 - a^{2}} \sqrt{1 - b^{2}} + a b

\Rightarrow A = {s i n}^{- 1} (\sqrt{(1 - a^{2}) (1 - b^{2})} + a b)

2) {s i n}^{- 1} a = x \Rightarrow a = s i n x \Rightarrow c o s x = \sqrt{1 - a^{2}}

{s i n}^{- 1} b = y \Rightarrow b = s i n y \Rightarrow c o s x = \sqrt{1 - b^{2}}

A = x + y \Rightarrow s i n A = s i n x c o s y + s i n y c o s x

\Rightarrow s i n A = a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}}

\Rightarrow A = {s i n}^{- 1} (a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}})

3) {c o s}^{- 1} a = x \Rightarrow a = c o s x \Rightarrow s i n x = \sqrt{1 - a^{2}}

{c o s}^{- 1} b = y \Rightarrow b = c o s y \Rightarrow s i n y = \sqrt{1 - b^{2}}

A = x + y \Rightarrow c o s A = c o s x c o s y - s i n x s i n y

c o s A = a b - \sqrt{1 - a^{2}} \sqrt{1 - b^{2}}

\Rightarrow A = {c o s}^{- 1} (a b - \sqrt{(1 - a^{2}) (1 - b^{2})})

Commented by pticantor last updated on 20/Aug/23

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