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1.)cos alpha+sin alpha*tan alpha=

2.) 1/(1+tan^2 alpha)=

3.)tan alpha/Wurzel(1+tan^2 alpha)=

4.)sin^4 alpha−cos^4 alpha=

 12.05.2015

Beste Antwort 

 #2
avatar+26387 
+5

4.) sin^4 alpha−cos^4 alpha=

$$\small{\text{$
\begin{array}{rcl}
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) } \\
&=& - (~ \cos^4{ ( \alpha ) } - \sin^4{ (\alpha) } ~)\\
&=& - (~
( \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } ) \cdot
( \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } )
~)\\\\
&&$Formeln:\\\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } = \cos{(2\cdot\alpha)} }}\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } = 1 }}\\\\
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) }&=&-\cos{(2\cdot\alpha)}
\end{array}
$}}$$

 13.05.2015
 #1
avatar+12531 
0

Omi67 12.05.2015
 #2
avatar+26387 
+5
Beste Antwort

4.) sin^4 alpha−cos^4 alpha=

$$\small{\text{$
\begin{array}{rcl}
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) } \\
&=& - (~ \cos^4{ ( \alpha ) } - \sin^4{ (\alpha) } ~)\\
&=& - (~
( \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } ) \cdot
( \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } )
~)\\\\
&&$Formeln:\\\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } - \sin^2{ (\alpha) } = \cos{(2\cdot\alpha)} }}\\
&&$~\boxed{\mathbf{ \cos^2{ ( \alpha ) } + \sin^2{ (\alpha) } = 1 }}\\\\
\sin^4{ (\alpha) } - \cos^4{ ( \alpha ) }&=&-\cos{(2\cdot\alpha)}
\end{array}
$}}$$

heureka 13.05.2015

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