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MishaVoro
@MishaVoro
August 2022
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Решите 3 задания. Тригонометрия
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ученик1484
1. a) cos 75 cos75=cos(45+30)=сos45*cos30-sin45*sin30=
= √2/2*√3/2-√2/2*1/2=√2/2(√3/2-1/2)≈0.35*0.73=0.25
b) = cos (67-7)=cos60=1/2
c) = sin(87+3)=sin90=1
2. cos2a=cos²a-sin²a=2cos²a-1
cos²a-cos2a=cos²a-2cos²a+1=1-cos²a
1-cos²a/(1-cos²a)=1
3. используем формулу тангенса разности
tg(α-β)=(tgα-tgβ)/(1+tgα*tgβ)
тогда получаем - заданное выражение = tg(5x-3x)=tg2x
tg2x=-√3 2x=2π/3+πk x=π/3+πk/2 k∈Z
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mikael2
Verified answer
Cos75=cos45cos30-sin45sin30=√2/2×√3/2-√2/2×1/2=(√6-√2)/4
sin67sin7+cos67cos7=cos(67-7)=cos60=1/2
sin87cos3+cos87sin3=sin(87+3)=sin90=1
(1-cos^2a)/(cos^2a-cos2a)=sin^2a/(cos^2a-cos^2a+sin^2a)=sin^2a/sin^2a=1
3)(tg5x-tg3x)/(1+tg5xtg3x)=
-√3
tg2x=-√3
2x=-пи/3+пиk
x=-пи/6+пи/2×k
0 votes
Thanks 0
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Answers & Comments
= √2/2*√3/2-√2/2*1/2=√2/2(√3/2-1/2)≈0.35*0.73=0.25
b) = cos (67-7)=cos60=1/2
c) = sin(87+3)=sin90=1
2. cos2a=cos²a-sin²a=2cos²a-1
cos²a-cos2a=cos²a-2cos²a+1=1-cos²a
1-cos²a/(1-cos²a)=1
3. используем формулу тангенса разности
tg(α-β)=(tgα-tgβ)/(1+tgα*tgβ)
тогда получаем - заданное выражение = tg(5x-3x)=tg2x
tg2x=-√3 2x=2π/3+πk x=π/3+πk/2 k∈Z
Verified answer
Cos75=cos45cos30-sin45sin30=√2/2×√3/2-√2/2×1/2=(√6-√2)/4sin67sin7+cos67cos7=cos(67-7)=cos60=1/2
sin87cos3+cos87sin3=sin(87+3)=sin90=1
(1-cos^2a)/(cos^2a-cos2a)=sin^2a/(cos^2a-cos^2a+sin^2a)=sin^2a/sin^2a=1
3)(tg5x-tg3x)/(1+tg5xtg3x)=
-√3
tg2x=-√3
2x=-пи/3+пиk
x=-пи/6+пи/2×k