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ilykee228
@ilykee228
June 2022
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Умоляю! Помогите с обратными тригонометрическими функциями!((
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LFP
Verified answer
1) ... = 3^3 = 27
2) --- = 3/4 = 0.75
3) =0
4) ... = 187/205
5) = pi - 4*(-pi / 2) = pi + 2pi = 3pi
1 votes
Thanks 3
ilykee228
Спасибо вам огромное!!! Вы бог!!!=)
LFP
нет, я человек... и иногда ошибаюсь)) надеюсь, не в этот раз...
ilykee228
хах) все мы ошибаемся) спасибо еще раз огромное!
oganesbagoyan
Verified answer
3arccosx +2arcsinx =3π/2⇔3(arccosx +arcsinx) - arcsinx = 3π/2⇔ 3π/2 - arcsinx = 3π/2
⇒x =0, следовательно (x+3)³ =(0+3)³
=27.
---
4arcsinx +arccosx =π ⇔3arcsinx+ (arcsinx +arccosx) =π ⇔3arcsinx +π/2 =π⇔
arcsinx =π/6 ⇒ x =1/2 ,следовательно 3x² =3*(1/2)²
=3/4
.
---
cos(arctq√3 +arccos√3/2) = cos(π/3 +π/6) = cosπ/2
=0.
---
cos(arcsin40/41 -arcsin4/5) =cos(arcsin40/41)*cos(arcsin4/5) +sin(arcsin40/41)*sin(arcsin4/5) =cos(arccos9/41)*cos(arccos3/5) +40/41*4/5 =
9/41*3/5) +40/41*4/5 =(27+160)/5*41
= 187/205
.
|a| ≤1 ,|b| ≤1 .
max(arccosa -4arccosb) = π -4(-π/2) = π +2π
=3π
.
0 votes
Thanks 1
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Answers & Comments
Verified answer
1) ... = 3^3 = 272) --- = 3/4 = 0.75
3) =0
4) ... = 187/205
5) = pi - 4*(-pi / 2) = pi + 2pi = 3pi
Verified answer
3arccosx +2arcsinx =3π/2⇔3(arccosx +arcsinx) - arcsinx = 3π/2⇔ 3π/2 - arcsinx = 3π/2⇒x =0, следовательно (x+3)³ =(0+3)³ =27.
---
4arcsinx +arccosx =π ⇔3arcsinx+ (arcsinx +arccosx) =π ⇔3arcsinx +π/2 =π⇔
arcsinx =π/6 ⇒ x =1/2 ,следовательно 3x² =3*(1/2)² =3/4.
---
cos(arctq√3 +arccos√3/2) = cos(π/3 +π/6) = cosπ/2 =0.
---
cos(arcsin40/41 -arcsin4/5) =cos(arcsin40/41)*cos(arcsin4/5) +sin(arcsin40/41)*sin(arcsin4/5) =cos(arccos9/41)*cos(arccos3/5) +40/41*4/5 =
9/41*3/5) +40/41*4/5 =(27+160)/5*41 = 187/205.
|a| ≤1 ,|b| ≤1 .
max(arccosa -4arccosb) = π -4(-π/2) = π +2π =3π.