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regishkam
@regishkam
August 2022
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Помогите решить пожалуйста :*
√2*tg a , если cos a = √6/3 и a лежит (П; 2П)
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karmelita1953
Sin²a=1-cos²a=1-(√6/3)²=1-6/9=3/9=1/3
sina=-√1/3=-√3/3
√2*tga=√2*sia/cosa=(-√2*√3/3)/ (√6/3=)=1
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Answers & Comments
sina=-√1/3=-√3/3
√2*tga=√2*sia/cosa=(-√2*√3/3)/ (√6/3=)=1