Tożsamości - dowody

[Z_i_o_M_e_K]

wykaż, że

a) \(\displaystyle{ (cos -cos ) ^{2} +(sin +sin ) ^{2} = 4sin ^{2} \frac{ - }{2}}\)

b) \(\displaystyle{ 2(1+cos )-sin ^{2} =4cos ^{4} \frac{ }{2}}\)

[soku11]

a)
\(\displaystyle{ (\cos\alpha-\cos\varphi)^2+(\sin\alpha+\sin\varphi)^2=4\sin^2\frac{\alpha-\varphi}{2}\\
=x\\
\varphi=y\\
(\cos x-\cos y)^2+(\sin x+\sin y)^2=4\sin^2\frac{x-y}{2}\\
L=
\cos^2x-2\cos x\cos y+\cos^2y+\sin^2x+2\sin x\sin y+\sin^2y=
\cos^2x+\sin^2x-2\cos x\cos y+\cos^2y+\sin^2 y+2\sin x\sin y=
2-2\cos x\cos y+2\sin x\sin y=
2-\cos(x+y)-\cos(x-y)+\cos(x+y)-\cos(x-y)=
2-2\cos(x-y)=
2[1-\cos(x-y)]=
2\cdot 2\sin^2 \frac{x-y}{2}=
4\sin^2 \frac{x-y}{2}=
4\sin^2 \frac{\alpha-\varphi}{2}=P}\)

[ Dodano: 21 Listopada 2008, 21:54 ]
b)
\(\displaystyle{ \alpha =x\\
\cos (2x)=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x\\
2(1+\cos x )-\sin^2 x=4\cos^4 \frac{x}{2}\\
L=
2\left(1+1-2\sin^2 \frac{x}{2}\right)-\sin^2 x=
4\left(1-\sin^2 \frac{x}{2}\right)-1+\cos^2x=
3-4\sin^2 \frac{x}{2}+\cos^2x=
3-4\sin^2 \frac{x}{2}+\left(1-2\sin^2 \frac{x}{2}\right)^2=
3-4\sin^2 \frac{x}{2}+1-4\sin^2 \frac{x}{2}+4\sin^4\frac{x}{2}=
4-8\sin^2 \frac{x}{2}+4\sin^4\frac{x}{2}=
4\left(\sin^4 \frac{x}{2}-2\sin ^2\frac{x}{2}+1\right)=
4\left( \sin^2 \frac{x}{2}-1\right)^2=
4\left( 1-\sin^2 \frac{x}{2}\right)^2=
4\left( \cos^2 \frac{x}{2}\right)^2=
4\cos^4 \frac{x}{2}=P}\)

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