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1. (1+sinx/1-sinx)·tg² (π/4-х/2)-sin² x
2. (1-2cos x/2 + cos x)/1+2cos x/2+cosx
3. (1-2sin x/2-cosx)/1+2sin x/2-cosx
1. (1+sinx/1-sinx)·tg² (π/4-х/2)-sin² x
2. (1-2cos x/2 + cos x)/1+2cos x/2+cosx
3. (1-2sin x/2-cosx)/1+2sin x/2-cosx
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Answers & Comments
1-sinx 4 2 (1-sinx)(1+sinx)
1.1) tg²(π - x) =sin²(π - x) : cos²(π - x)=0.5(1-sinx)= 1-sinx
4 2 4 2 4 2 0.5(1+sinx) 1+sinx
1.1.1) sin²(π -x)=(sin π cos x - cos π sin x )²=(√2 cos x - √2 sin x)²=
4 2 4 2 4 2 2 2
=(√2(cos x - sin x))²= 2 (cos² x - 2 cos x sin x + sin² x) =
2 2 2 4 2 2 2 2
=0.5 (1-sin 2*x)=0.5 (1-sin x)
2
1.1.2) cos²(π - x)=(cos π cos x + sin π sin x)²=(√2 cos x + √2 sin x)²=
4 2 4 2 4 2 2 2 2 2
= (√2(cos x + sin x))²=2 (cos² x + 2 cos x sin x + sin² x)=
2 2 2 4 2 2 2 2
=0.5 (1+sin x)
2) 1-2 cos (x/2) + cos x =1-2cos(x/2)+2cos²(x/2)-1=2cos(x/2)(cos(x/2)-1) =
1+2 cos(x/2) + cos x 1+2cos(x/2)+2cos²(x/2)-1 2cos(x/2)(cos(x/2)+1)
2.1) cos x=cos (2*(x/2))=cos² (x/2) - sin² (x/2)=cos²(x/2)-(1-cos²(x/2))=
=cos²(x/2)-1+cos²(x/2)=2cos²(x/2)-1
=cos(x/2)-1
cos(x/2)+1
3) 1-2 sin(x/2) - cosx=1-2sin(x/2)-1+2sin²(x/2)= 2sin(x/2)(sin(x/2)-1)=
1+2sin(x/2)- cosx 1+2sin(x/2)-1+2sin²(x/2) 2sin(x/2)(sin(x/2)+1)
3.1) cos x= cos²(x/2) - sin²(x/2)=1-sin²(x/2)-sin²(x/2)=1-2sin²(x/2)
=sin(x/2)-1
sin(x/2)+1