Ответ:
3/2
Пошаговое объяснение:
tg(a-b)=(tga-tgb)/(1+tga·tgb)
tg(3π/4)= -1
sin²α+cos²α=1
sin2α=2·sinα·cosα
tg(3π/4-α)=(tg3π/4-tgα)/(1+tg3π/4·tgα)=(-1-tgα)/(1+(-1)·tgα)=
= -(1+tgα)/(1-tgα)=-(cosα+sinα)/(cosα-sinα)
tg²(3π/4-α)=(-(cosα+sinα)/(cosα-sinα))²=
=(cos²α+2·cosα·sinα+sin²α)/(cos²α-2·cosα·sinα+sin²α)=(1+sin2α)/(1-sin2α)
3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)
sin2α=-1/3
3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)=3·(1+(-1/3))/(1-(-1/3))=3·(2/3)/(1+1/3)=
=3·(2/3)/(4/3)=3·(2/3)·(3/4)=3/2
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Verified answer
Ответ:
3/2
Пошаговое объяснение:
tg(a-b)=(tga-tgb)/(1+tga·tgb)
tg(3π/4)= -1
sin²α+cos²α=1
sin2α=2·sinα·cosα
tg(3π/4-α)=(tg3π/4-tgα)/(1+tg3π/4·tgα)=(-1-tgα)/(1+(-1)·tgα)=
= -(1+tgα)/(1-tgα)=-(cosα+sinα)/(cosα-sinα)
tg²(3π/4-α)=(-(cosα+sinα)/(cosα-sinα))²=
=(cos²α+2·cosα·sinα+sin²α)/(cos²α-2·cosα·sinα+sin²α)=(1+sin2α)/(1-sin2α)
3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)
sin2α=-1/3
3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)=3·(1+(-1/3))/(1-(-1/3))=3·(2/3)/(1+1/3)=
=3·(2/3)/(4/3)=3·(2/3)·(3/4)=3/2