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bangbang635
@bangbang635
August 2022
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1+sinα+cosα=2[tex] \sqrt{2} [/tex] * cos α/2*cos(π/4-α/2)
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sedinalana
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2√2*1/2*(cos(a/2-π/4+a/2)+cos(a/2+π/4-a/2))=
=√2(cos(a-π/4)+cosπ/4)=√2*(cosacosπ/4+sinasinπ/4)+√2*√2/2=
=√2*(√2/2*cosa+√/2*sina)+1=√2*√2/2(cosa+sina)+1=
=cosa+sina+1
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Answers & Comments
Verified answer
2√2*1/2*(cos(a/2-π/4+a/2)+cos(a/2+π/4-a/2))==√2(cos(a-π/4)+cosπ/4)=√2*(cosacosπ/4+sinasinπ/4)+√2*√2/2=
=√2*(√2/2*cosa+√/2*sina)+1=√2*√2/2(cosa+sina)+1=
=cosa+sina+1