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Question Number 160746 by Rokon last updated on 05/Dec/21

Answered by som(math1967) last updated on 06/Dec/21

if ๐›‰=45ยฐthen A=(โˆš2) B=0  ii) A=(โˆš2)(cosฮธ+sin๐›‰โˆ’sin๐›‰)  โˆดcos๐›‰+sin๐›‰=(โˆš2)cos๐›‰  โ‡’sin๐›‰=((โˆš2)โˆ’1)cos๐›‰  โ‡’((โˆš2)+1)sin๐›‰=((โˆš2)+1)((โˆš2)โˆ’1)cos๐›‰  โ‡’((โˆš2)+1)sin๐›‰=cos๐›‰  โ‡’(โˆš2)sin๐›‰=cos๐›‰โˆ’sin๐›‰  โ‡’(โˆš2)(cos๐›‰+sin๐›‰โˆ’cos๐›‰)=B  โˆดB=(โˆš2)(Aโˆ’cos๐›‰)

$${if}\:\boldsymbol{\theta}=\mathrm{45}ยฐ\boldsymbol{{then}}\:\boldsymbol{{A}}=\sqrt{\mathrm{2}}\:\boldsymbol{{B}}=\mathrm{0} \\ $$$$\left.{ii}\right)\:{A}=\sqrt{\mathrm{2}}\left({cos}\theta+\boldsymbol{{sin}\theta}โˆ’{sin}\boldsymbol{\theta}\right) \\ $$$$\therefore\boldsymbol{{cos}\theta}+\boldsymbol{{sin}\theta}=\sqrt{\mathrm{2}}\boldsymbol{{cos}\theta} \\ $$$$\Rightarrow\boldsymbol{{sin}\theta}=\left(\sqrt{\mathrm{2}}โˆ’\mathrm{1}\right)\boldsymbol{{cos}\theta} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\boldsymbol{{sin}\theta}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}โˆ’\mathrm{1}\right)\boldsymbol{{cos}\theta} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\boldsymbol{{sin}\theta}=\boldsymbol{{cos}\theta} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\boldsymbol{{sin}\theta}=\boldsymbol{{cos}\theta}โˆ’\boldsymbol{{sin}\theta} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\left({c}\boldsymbol{{os}\theta}+\boldsymbol{{sin}\theta}โˆ’\boldsymbol{{cos}\theta}\right)=\boldsymbol{{B}} \\ $$$$\therefore\boldsymbol{{B}}=\sqrt{\mathrm{2}}\left(\boldsymbol{{A}}โˆ’\boldsymbol{{cos}\theta}\right) \\ $$

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