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Makast
@Makast
July 2022
1
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Упростить выражение и решить систему неравенств
[tex]sin^{2} ( \frac{9 \pi }{8} + \alpha )-sin^{2} ( \frac{17 \pi }{8} - \alpha )[/tex]
[tex] \left \{ {{ y^{2}+xy+y=28 } \atop {x^{2}+xy+x= 14}} \right. [/tex]
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sedinalana
Verified answer
1
sin²(π+π/8+a)-sin²(2π+π/8-a)=sin²(π/8+a)-sin²(π/8-a)=
=[sin(π/8+a)-sin(π/8-a)]*[sin(π/8+a)+sin(π/8-a)]=
=2sina*cosπ/8*2sinπ/8*cosa=sin2a*sinπ/4=√2/2*sin2a
2
{y²+xy+y=28
{x²+xy+x=14
отнимем
y²-x²+y-x=14
(y-x)(y+x)+(y-x)=14
(y-x)(y+x+1)=14
1){y-x=2
{x+y+1=7
2y+1=9
2y=8
y=4
4-x=2
x=2
2){y-x=7
{y+x+1=2
2y+1=9
2y=8
y=4
4-x=7
x=-3
3)y-x=-2
{y+x+1=-7
2y+1=-9
2y=-10
y=-5
-5-x=-2
x=-3
4){y-x=-7
{y+x+1=-2
2y+1=-9
2y=-10
y=-5
-5-x=-7
x=2
проверка
(2;4)
{16+8+4=28 28=28
{4+8+4=14 14=14
(-3;4)
{16-12+4=8 8≠28
{9-12-3=-6 -6≠14
(-3;-5)
25+15-5 25≠28
{9+15-3=21 21≠14
(2;-5)
{25-10-5=10 10≠28
{4-10+2=-4 -4≠14
Ответ (2;4)
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Answers & Comments
Verified answer
1sin²(π+π/8+a)-sin²(2π+π/8-a)=sin²(π/8+a)-sin²(π/8-a)=
=[sin(π/8+a)-sin(π/8-a)]*[sin(π/8+a)+sin(π/8-a)]=
=2sina*cosπ/8*2sinπ/8*cosa=sin2a*sinπ/4=√2/2*sin2a
2
{y²+xy+y=28
{x²+xy+x=14
отнимем
y²-x²+y-x=14
(y-x)(y+x)+(y-x)=14
(y-x)(y+x+1)=14
1){y-x=2
{x+y+1=7
2y+1=9
2y=8
y=4
4-x=2
x=2
2){y-x=7
{y+x+1=2
2y+1=9
2y=8
y=4
4-x=7
x=-3
3)y-x=-2
{y+x+1=-7
2y+1=-9
2y=-10
y=-5
-5-x=-2
x=-3
4){y-x=-7
{y+x+1=-2
2y+1=-9
2y=-10
y=-5
-5-x=-7
x=2
проверка
(2;4)
{16+8+4=28 28=28
{4+8+4=14 14=14
(-3;4)
{16-12+4=8 8≠28
{9-12-3=-6 -6≠14
(-3;-5)
25+15-5 25≠28
{9+15-3=21 21≠14
(2;-5)
{25-10-5=10 10≠28
{4-10+2=-4 -4≠14
Ответ (2;4)