MHB Prove: $(\sin \theta+ i \cos \theta)^8 = \cos 8\theta - i \sin 8\theta$

[kaliprasad]

Prove that $(\sin \theta+ i \cos \theta)^8 = \cos 8\theta - i \sin 8\theta$

 

[Greg]

Prove that $(\sin \theta+ i \cos \theta)^8 = \cos 8\theta - i \sin 8\theta$

Spoiler

Using Euler's formula,

$$\begin{align*}(\sin\theta+i\cos\theta)^8&=\left(\cos\left(\dfrac{\pi}{2}-\theta\right)+i\sin\left(\dfrac{\pi}{2}-\theta\right)\right)^8 \\
&=\left(e^{\left(\dfrac{\pi}{2}-\theta\right)i}\right)^8 \\
&=e^{(4\pi-8\theta)i} \\
&=\cos8\theta-i\sin8\theta\end{align*}$$

 

[kaliprasad]

Spoiler

Using Euler's formula,

$$\begin{align*}(\sin\theta+i\cos\theta)^8&=\left(\cos\left(\dfrac{\pi}{2}-\theta\right)+i\sin\left(\dfrac{\pi}{2}-\theta\right)\right)^8 \\
&=\left(e^{\left(\dfrac{\pi}{2}-\theta\right)i}\right)^8 \\
&=e^{(4\pi-8\theta)i} \\
&=\cos8\theta-i\sin8\theta\end{align*}$$

Spoiler

nice

 

[kaliprasad]

Spoiler

Using Euler's formula,

$$\begin{align*}(\sin\theta+i\cos\theta)^8&=\left(\cos\left(\dfrac{\pi}{2}-\theta\right)+i\sin\left(\dfrac{\pi}{2}-\theta\right)\right)^8 \\
&=\left(e^{\left(\dfrac{\pi}{2}-\theta\right)i}\right)^8 \\
&=e^{(4\pi-8\theta)i} \\
&=\cos8\theta-i\sin8\theta\end{align*}$$

above is a good ans
my answer different from above is as below

Spoiler

Using Euler's formula,
$(\sin\theta+i\cos\theta)^8 = i^8(\cos\theta - i \sin \theta)^8$
=$(e^{-i\theta})^8= e^{-i8\theta}$
= $\cos 8\theta - i \sin 8\theta$

 

[Prove It]

Prove that $(\sin \theta+ i \cos \theta)^8 = \cos 8\theta - i \sin 8\theta$

Spoiler

$\displaystyle \begin{align*} \left[ \sin{ \left( \theta \right) } + \mathrm{i}\cos{ \left( \theta \right) } \right] ^8 &= \left\{ \mathrm{i}\,\left[ \cos{ \left( \theta \right) } - \mathrm{i }\sin{\left( \theta \right) } \right] \right\} ^8 \\ &= \mathrm{i}^8\,\left[ \cos{ \left( \theta \right) } - \mathrm{i}\sin{ \left( \theta \right) } \right] ^8 \\ &= 1\,\left[ \cos{ \left( \theta \right) } - \mathrm{i}\sin{ \left( \theta \right) } \right] ^8 \\ &= \left[ \cos{ \left( \theta \right) } - \mathrm{i}\sin{\left( \theta \right) } \right] ^8 \\ &= \left[ \cos{ \left( -\theta \right) } + \mathrm{i}\sin{ \left( -\theta \right) } \right] ^8 \\ &= \left( \mathrm{e}^{-\mathrm{i}\,\theta} \right) ^8 \\ &= \mathrm{e}^{ -8\,\mathrm{i}\,\theta } \\ &= \cos{ \left( -8\,\theta \right) } + \mathrm{i}\sin{ \left( -8\,\theta \right) } \\ &= \cos{ \left( 8\,\theta \right) } - \mathrm{i}\sin{ \left( 8\,\theta \right) } \end{align*}$