# CS代写|计算机图形学作业代写computer graphics代考|CSCl2240

## CS代写|计算机图形学作业代写computer graphics代考|Double-Angle Identities

By making $\beta=\alpha$, the three compound-angle identities
\begin{aligned} \sin (\alpha \pm \beta) &=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \ \cos (\alpha \pm \beta) &=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \ \tan (\alpha \pm \beta) &=\frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} \end{aligned}
provide the starting point for deriving three corresponding double-angle identities:
\begin{aligned} \sin (\alpha \pm \alpha) &=\sin \alpha \cos \alpha \pm \cos \alpha \sin \alpha \ \sin (2 \alpha) &=2 \sin \alpha \cos \alpha . \end{aligned}
Similarly,
\begin{aligned} \cos (\alpha \pm \alpha) &=\cos \alpha \cos \alpha \mp \sin \alpha \sin \alpha \ \cos (2 \alpha) &=\cos ^2 \alpha-\sin ^2 \alpha \end{aligned}
which can be further simplified using $\sin ^2 \alpha+\cos ^2 \alpha=1$ :

\begin{aligned} &\cos (2 \alpha)=\cos ^2 \alpha-\sin ^2 \alpha \ &\cos (2 \alpha)=2 \cos ^2 \alpha-1 \ &\cos (2 \alpha)=1-2 \sin ^2 \alpha . \end{aligned}
And for $\tan (2 \alpha)$, we have:
\begin{aligned} \tan (\alpha \pm \alpha) &=\frac{\tan \alpha \pm \tan \alpha}{1+\tan \alpha \tan \alpha} \ \tan (2 \alpha) &=\frac{2 \tan \alpha}{1-\tan ^2 \alpha} . \end{aligned}

## CS代写|计算机图形学作业代写computer graphics代考|Area of a Shape

The area of a polygonal shape is readily calculated from its list of coordinates. For example, using the list of coordinates shown in Table 5.1: the area is computed by $$\text { area }=\frac{1}{2}\left[\left(x_0 y_1-x_1 y_0\right)+\left(x_1 y_2-x_2 y_1\right)+\left(x_2 y_3-x_3 y_2\right)+\left(x_3 y_0-x_0 y_3\right)\right] \text {. }$$
You will observe that the calculation sums the results of multiplying an $x$ by the next $y$, minus the next $x$ by the previous $y$. When the last vertex is selected, it is paired with the first vertex to complete the process. The result is then halved to reveal the area. As a simple test, let’s apply this formula to the shape descrihed in Fig. 5.3:
\begin{aligned} &\text { area }=\frac{1}{2}[(1 \times 1-3 \times 1)+(3 \times 2-3 \times 1)+(3 \times 3-1 \times 2)+(1 \times 1-1 \times 3)] \ &\text { area }=\frac{1}{2}[-2+3+7-2]=3 . \end{aligned}
which, by inspection, is the true area. The beauty of this technique is that it works with any number of vertices and any arbitrary shape. The origin of this technique is revealed in Chap. 7.

Another feature of the technique is that if the set of coordinates is clockwise, the area is negative, which means that the calculation computes vertex orientation as well as area. To illustrate this feature, the original vertices are reversed to a clockwise sequence as follows:
\begin{aligned} &\text { area }=\frac{1}{2}[(1 \times 3-1 \times 1)+(1 \times 2-3 \times 3)+(3 \times 1-3 \times 2)+(3 \times 1-1 \times 1)] \ &\text { area }=\frac{1}{2}[2-7-3+2]=-3 . \end{aligned}
The minus sign confirms that the vertices are in a clockwise sequence.

# 计算机图形学代考

## CS代写|计算机图形学作业代写computer graphics代考|Double-Angle Identities

$$\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \cos (\alpha \pm \beta) \quad=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \tan (\alpha \pm \beta)=\frac{\tan \alpha \pm}{1 \mp \tan \alpha}$$

$$\sin (\alpha \pm \alpha)=\sin \alpha \cos \alpha \pm \cos \alpha \sin \alpha \sin (2 \alpha) \quad=2 \sin \alpha \cos \alpha .$$

$$\cos (\alpha \pm \alpha)=\cos \alpha \cos \alpha \mp \sin \alpha \sin \alpha \cos (2 \alpha) \quad=\cos ^2 \alpha-\sin ^2 \alpha$$

$$\cos (2 \alpha)=\cos ^2 \alpha-\sin ^2 \alpha \quad \cos (2 \alpha)=2 \cos ^2 \alpha-1 \cos (2 \alpha)=1-2 \sin ^2 \alpha .$$

$$\tan (\alpha \pm \alpha)=\frac{\tan \alpha \pm \tan \alpha}{1+\tan \alpha \tan \alpha} \tan (2 \alpha) \quad=\frac{2 \tan \alpha}{1-\tan ^2 \alpha}$$

## CS代写|计算机图形学作业代写computer graphics代考|Area of a Shape

$$\text { area }=\frac{1}{2}\left[\left(x_0 y_1-x_1 y_0\right)+\left(x_1 y_2-x_2 y_1\right)+\left(x_2 y_3-x_3 y_2\right)+\left(x_3 y_0-x_0 y_3\right)\right] .$$

$$\text { area }=\frac{1}{2}[(1 \times 1-3 \times 1)+(3 \times 2-3 \times 1)+(3 \times 3-1 \times 2)+(1 \times 1-1 \times 3)] \quad \text { area }=\frac{1}{2}[-2+$$

$$\text { area }=\frac{1}{2}[(1 \times 3-1 \times 1)+(1 \times 2-3 \times 3)+(3 \times 1-3 \times 2)+(3 \times 1-1 \times 1)] \quad \text { area }=\frac{1}{2}[2-7$$

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