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Question Number 196166 by pticantor last updated on 19/Aug/23
for any a,b∈[−1,1] calculate: arccos(a)+arcsin(b)=? arcsin(a)+arcsin(b)=? arccos(a)+arccos(b)=?
$$\boldsymbol{{for}}\:\boldsymbol{{any}}\:\boldsymbol{{a}},\boldsymbol{{b}}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\boldsymbol{{calculate}}:\: \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arcsin}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arc}}{cos}\left(\boldsymbol{{b}}\right)=? \\ $$
Commented by pticantor last updated on 19/Aug/23
i make some unforgotable mistake! please
$${i}\:{make}\:{some}\:{unforgotable}\:{mistake}!\:{please}\: \\ $$
Answered by MM42 last updated on 19/Aug/23
1)cos^(−1) a=x⇒a=cosx⇒sinx=(√(1−a^2 )) sin^(−1) b=y⇒b=siny⇒cosy=(√(1−b^2 )) A=x+y⇒sinA=sinxcosy+sinycosx ⇒sinA=(√(1−a^2 ))(√(1−b^2 ))+ab ⇒A=sin^(−1) ((√((1−a^2 )(1−b^2 )))+ab) 2)sin^(−1) a=x⇒a=sinx⇒cosx=(√(1−a^2 )) sin^(−1) b=y⇒b=siny⇒cosx=(√(1−b^2 )) A=x+y⇒sinA=sinxcosy+sinycosx ⇒sinA=a(√(1−b^2 ))+b(√(1−a^2 )) ⇒A=sin^(−1) (a(√(1−b^2 ))+b(√(1−a^2 ))) 3)cos^(−1) a=x⇒a=cosx⇒sinx=(√(1−a^2 )) cos^(−1) b=y⇒b=cosy⇒siny=(√(1−b^2 )) A=x+y⇒cosA=cosxcosy−sinxsiny cosA=ab−(√(1−a^2 ))(√(1−b^2 )) ⇒A=cos^(−1) (ab−(√((1−a^2 )(1−b^2 ) )))
$$\left.\mathrm{1}\right){cos}^{−\mathrm{1}} {a}={x}\Rightarrow{a}={cosx}\Rightarrow{sinx}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${sin}^{−\mathrm{1}} {b}={y}\Rightarrow{b}={siny}\Rightarrow{cosy}=\sqrt{\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$${A}={x}+{y}\Rightarrow{sinA}={sinxcosy}+{sinycosx} \\ $$$$\Rightarrow{sinA}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{ab} \\ $$$$\Rightarrow{A}={sin}^{−\mathrm{1}} \left(\sqrt{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\left(\mathrm{1}−{b}^{\mathrm{2}} \right)}+{ab}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right){sin}^{−\mathrm{1}} {a}={x}\Rightarrow{a}={sinx}\Rightarrow{cosx}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${sin}^{−\mathrm{1}} {b}={y}\Rightarrow{b}={siny}\Rightarrow{cosx}=\sqrt{\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$${A}={x}+{y}\Rightarrow{sinA}={sinxcosy}+{sinycosx} \\ $$$$\Rightarrow{sinA}={a}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}={sin}^{−\mathrm{1}} \left({a}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\left.\mathrm{3}\right){cos}^{−\mathrm{1}} {a}={x}\Rightarrow{a}={cosx}\Rightarrow{sinx}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${cos}^{−\mathrm{1}} {b}={y}\Rightarrow{b}={cosy}\Rightarrow{siny}=\sqrt{\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$${A}={x}+{y}\Rightarrow{cosA}={cosxcosy}−{sinxsiny} \\ $$$${cosA}={ab}−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\sqrt{\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}={cos}^{−\mathrm{1}} \left({ab}−\sqrt{\left(\mathrm{1}−{a}^{\mathrm{2}} \right)\left(\mathrm{1}−{b}^{\mathrm{2}} \right)\:}\right) \\ $$$$ \\ $$
Commented by pticantor last updated on 20/Aug/23
thank you too much sir
$${thank}\:{you}\:{too}\:{much}\:{sir} \\ $$

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