Решите пожалуйста вариант 8, 5 заданий. Created by beatles1970. algebra-ru

Решите пожалуйста вариант 8, 5 заданий...

beatles1970 @beatles1970

August 2022 1 5 Report

Решите пожалуйста вариант 8, 5 заданий

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oganesbagoyan

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1) 88 sin²α -90cos²α =10.
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ctq²α -?
88 sin²α -90cos²α =10 ;
88 sin²α -90cos²α =10(sin²α+cos²α) ;
88 sin²α -90cos²α =10sin²α+10cos²α ;
78 sin²α=100 cos²α ;
sin²α=(50/39) *cos²α ;
ctq²α =50/39.

ответ: 50/39.
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2) tqα =√5 ; α∈(π/2 ;3π/2) .
---------------
√30sinα - ?
по условию  tqα =√5 >0 ⇒α∈(π;3π/2) , где sinα <0.
ctqα =1/tqα =1 / √5 ;
1+ctq²α =1/sin²α ;:
sinα = - 1/√(1+ctq²α) .
sinα = -1/√(1+1/5) = -√5/√6 ;
√30sinα =√30*(-√5/√6)= -√5*√6*√5/√6) = - 5 .

ответ: -5.
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3) tq600°*sin1200°*cos2400° =tq(2*360° -120°)* sin(3*360°+120°) *cos(7*360°-120°) =
= - tq120°*sin120°*cos120° = -sin²120° = -sin²60° = -3/4.

ответ: 0,75.
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4) (os²23π/16-cos²33π/16)*sin33π/8*sin33π/4 .
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(cos²23π/16-cos²33π/16)sin33π/8*sin33π/4  =    ( (1+cos23π/8 -1-cos33π/8)/2))sin(4π +π/8)sin(8π+π/4) = ((cos23π/8 -cos33π/8 )/2)sinπ/8*sinπ/4=
sin5π/8 *sin7π/2 *sinπ/8*sinπ/4= - sin5π/8 *sinπ/8*sinπ/4= - sin(π/2+π/8)sinπ/8*sinπ/4 =
-(cosπ/8* sinπ/8)*sinπ/4 = -0,5*sinπ/4*sinπ/4 = -0,5sin²π/4 = -0,5*0,5 = -0,25.

ответ: - 0,25.
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5) √6(sin25°cos85° +sin245°sin85°)/(sin20°sin65°- sin290°cos65°) =
√6(sin25°sin5° -cos25°cos5°)/(sin20°cos25°+ cos20°sins25°) =
-√6(cos25°cos5° - sin25°sin5°)/sin(20°+25°)=
-√6cos(25°+5°)/sin45°= -√6 cos30°*(√2)/2 = -√3 *(√3)/2 = -1,5 .

ответ: -1,5.

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