Удвоени ъгли - помощ

[Гост]

Докажете ,че

sin[tex]^{4 }[/tex]d+ cos[tex]^{4 }[/tex]d = [tex]\frac{1}{2}[/tex](1+ cos[tex]^{2 }[/tex]2d)

sin[tex]^{6 }[/tex]d + cos[tex]^{6 }[/tex]d= [tex]\frac{1}{4}[/tex]
(1+3 cos[tex]^{2 }[/tex]2d)

[tex]\frac{cos4d.tg2d-sin4d}{cos4d.cotg2d+sin4d}[/tex]=-tg[tex]^{2 }[/tex]2d

[ammornil]

Докажете ,че

sin[tex]^{4 }[/tex]d+ cos[tex]^{4 }[/tex]d = [tex]\frac{1}{2}[/tex](1+ cos[tex]^{2 }[/tex]2d)

sin[tex]^{6 }[/tex]d + cos[tex]^{6 }[/tex]d= [tex]\frac{1}{4}[/tex]
(1+3 cos[tex]^{2 }[/tex]2d)

[tex]\frac{cos4d.tg2d-sin4d}{cos4d.cotg2d+sin4d}[/tex]=-tg[tex]^{2 }[/tex]2d

$\\[12pt] a^{4}+b^{4}= (a^{2} +b^{2})^{2} -2ab \Rightarrow \\[6pt] \sin^{4}{d} +\cos^{4}{d}= (\sin^{2}{d} +\cos^{2}{d})^{2} -2\sin^{2}{d}\cos^{2}{d} =1^{2} -\dfrac{1}{2}\cdot{}4\sin^{2}{d}\cos^{2}{d} = \\[6pt] = 1- \dfrac{1}{2}(2\sin{d}\cos{d})^{2}= 1 -\dfrac{1}{2}\sin^{2}{2d}= \dfrac{1}{2}[2 -1 +\cos^{2}{2d})= \dfrac{1}{2}(1+\cos^{2}{2d}) \\[24pt] \begin{array}{l} \hline \hspace{12em} \end{array} \\[6pt] \sin^{6}{d} + \cos^{6}{d}= \frac{1}{4}(1+3 \cos^{2}{2d}) \\[6pt] a^{6}+b^{6} =(a^{2})^{3} +b(^{2})^{3}= (a^{2} +b^{2})^{3} -3a^{2}b^{2}(a^{2}+b^{2}), \quad a=\sin{d}, \quad b=\cos{d} \Rightarrow \cdots{}\\[24pt] \begin{array}{l} \hline \hspace{12em} \end{array} \\[6pt] \dfrac{\cos{4d}\cdot{}\tg{2d}-\sin{4d}}{\cos{4d}\cdot{}\cotg{2d}+\sin{4d}}=-\tg^{2}{2d} \\[6pt] \text{ДМ}\quad \begin{array}{|l} 2d \ne{} \pm\dfrac{\pi}{2} \pm{k\pi} \\ 2d \ne{} 0 \pm{k\pi} \end{array} \Rightarrow \begin{array}{|l} d \ne{} \pm\dfrac{\pi}{4} \pm{k\dfrac{\pi}{2}} \\ d \ne{} 0 \pm{k\dfrac{\pi}{2}} \end{array}, \hspace{0.2em} k \in\mathbb{N_{0}} \\[12pt] \sin{4d}=\dfrac{2\tg{2d}}{1+\tg^{2}{2d}}, \quad \cos{4d}=\dfrac{1-\tg^{2}{2d}}{1+\tg^{2}{2d}} \\[6pt] \because{}\tg{2d}= t \Rightarrow \dfrac{\cos{4d}\cdot{}\tg{2d}-\sin{4d}}{\cos{4d}\cdot{}\cotg{2d}+\sin{4d}}= \dfrac{\dfrac{1-t^{2}}{1+t^{2}}t-\dfrac{2t}{1+t^{2}}}{\quad \dfrac{1-t^{2}}{1+t^{2}}\dfrac{1}{t}+\dfrac{2t}{1+t^{2}} \quad}= \dfrac{\dfrac{t(1-t^{2})-2t}{1+t^{2}}}{\quad \dfrac{1-t^{2} +2t^{2}}{t(1+t^{2})} \quad}= \\[6pt] = t\cdot{}\dfrac{t(1-t^{2})-2t}{1+t^{2}}= \dfrac{t^{2} -t^{4} -2t^{2}}{1+t^{2}}=\dfrac{-t^{4}-t^{2}}{1+t^{2}}= \dfrac{-t^{2}(t^{2}+1)}{t^{2}+1}=-t^{2} =-\tg^{2}{2d}$

[S.B.]

Докажете ,че 4
sin[tex]^{4 }[/tex]d+ cos[tex]^{4 }[/tex]d = [tex]\frac{1}{2}[/tex](1+ cos[tex]^{2 }[/tex]2d)

sin[tex]^{6 }[/tex]d + cos[tex]^{6 }[/tex]d= [tex]\frac{1}{4}[/tex]
(1+3 cos[tex]^{2 }[/tex]2d)

[tex]\frac{cos4d.tg2d-sin4d}{cos4d.cotg2d+sin4d}[/tex]=-tg[tex]^{2 }[/tex]2d

Още един поглед върху задачите:

1)
Нека [tex]A = \sin^{4 } \alpha + \cos^{4 } \alpha , B = \frac{1}{2} (1 + \cos^{2 }2 \alpha)[/tex]
Ще докажем $A = B$

[tex]A = \sin^{4 } \alpha + \cos^{2 } \alpha = ( \sin^{2 } \alpha ) ^{2 } + ( \cos^{2 } \alpha ) ^{2 } = ( \sin^{2 } \alpha + \cos^{2 } \alpha) ^{2 } - 2 \sin^{2 } \alpha \cos^{2 } \alpha =[/tex]

[tex]= 1 - \frac{1}{2} \sin^{2 }2 \alpha = 1 - \frac{1}{2}(1 - \cos^{2 }2 \alpha ) = 1 - \frac{1}{2}+ \frac{1}{2} \cos^{2 }2 \alpha = \frac{1}{2} + \frac{1}{2} \cos^{2 }2 \alpha =[/tex]

[tex]= \frac{1}{2}(1 + \cos^{2 }2 \alpha ) = B[/tex]
$$\Rightarrow A = B$$

2)
Нека [tex]A = \sin^{6 } \alpha + \cos^{6 } \alpha , B = \frac{1}{4}(1 + 3 \cos^{2 }2 \alpha)[/tex]
Ще докажем $A = B$

[tex]A = \sin^{6 } \alpha + \cos^{6 } \alpha = ( \sin^{2 } \alpha ) ^{3 } + ( \cos^{2 } \alpha )^{3 } =[/tex]

[tex]= ( \sin^{2 } \alpha + \cos^{2 } \alpha)( \sin^{4 } \alpha+ \cos^{4 } \alpha - \sin^{2 } \alpha \cos^{2 } \alpha) =[/tex]

Ще използвам резултата от предишната задача

[tex]= 1.[ \frac{1}{2}(1 + \cos^{2 }2 \alpha) - \frac{1}{4} \sin^{2 }2 \alpha ] = \frac{1}{2} + \frac{1}{2} \cos^{2 }2 \alpha - \frac{1}{4} \sin^{2 }2 \alpha = \frac{1}{2} + \frac{1}{2} \cos^{2 }2 \alpha + \frac{1}{4}(1 - \cos^{2 }2 \alpha) =[/tex]

[tex]= \frac{1}{2} + \frac{1}{2} \cos^{2 }2 \alpha - \frac{1}{4} + \frac{1}{4} \cos^{2 }2 \alpha = \frac{1}{4} + \frac{3}{4} \cos^{2 }2 \alpha = \frac{1}{4}(1 + 3 \cos^{2 }2 \alpha) = B[/tex]
$$\Rightarrow A = B$$

3)
Нека [tex]A = \frac{\cos 4 \alpha .\tg 2 \alpha - \sin 4 \alpha }{\cos 4 \alpha.\cotg 2 \alpha + \sin 4 \alpha } , B = - tg^{2 }2 \alpha[/tex]
Ще докажем $A=B$

[tex]A = \displaystyle \frac{\cos 4 \alpha.tg 2 \alpha - \sin 4 \alpha }{\cos 4 \alpha.\cotg 2 \alpha - \sin 4 \alpha } = \displaystyle \frac{\cos 4 \alpha .\displaystyle \frac{\sin 2 \alpha }{\cos 2 \alpha }- \sin 4 \alpha }{\cos 4 \alpha.\displaystyle \frac{\cos 2 \alpha }{\sin 2 \alpha } + \sin 4 \alpha } = \displaystyle \frac{\displaystyle \frac{\cos 4 \alpha\sin 2 \alpha - \sin 4 \alpha \cos 2 \alpha }{\cos 2 \alpha } }{\displaystyle \frac{\cos 4 \alpha\cos 2 \alpha + \sin 4 \alpha \sin 2 \alpha }{\sin 2 \alpha } } =[/tex]

[tex]= \frac{\sin (-2 \alpha)\sin2 \alpha }{\cos 2 \alpha\cos 2 \alpha } = \frac{- \sin^{2 }2 \alpha }{ \cos^{2 }2 \alpha } = - tg^{2 }2 \alpha = B[/tex]
$$\Rightarrow A = B$$

[Гост]

Много ви благодаря за помощта . Вече разбрах как да ги решавам.