# 计算机代写|计算机图形学代写computer graphics代考|COS426

## 计算机代写|计算机图形学代写computer graphics代考|Compound-Angle Identities

Trigonometric identities are useful for solving various mathematical problems, but apart from this, their proof often contains a strategy that can be used else where. In the first example, watch out for the technique of multiplying by 1 in the form of a ratio, and swapping denominators. The technique is rather elegant and suggests that the result was known in advance, which probably was the case. Let’s begin by finding a way of representing $\sin (\alpha+\beta)$ in terms of $\sin \alpha, \cos \alpha, \sin \beta, \cos \beta$.
With reference to $\mathrm{Fig} .4 .12$ :
\begin{aligned} \sin (\alpha+\beta) &=\frac{F D}{A D}=\frac{B C+E D}{A D} \ &=\frac{B C}{A D} \frac{A C}{A C}+\frac{E D}{A D} \frac{C D}{C D} \ &=\frac{B C}{A C} \frac{A C}{A D}+\frac{E D}{C D} \frac{C D}{A D} \ \sin (\alpha+\beta) &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}
To find $\sin (\alpha-\beta)$, reverse the sign of $\beta$ in (4.1):
$$\sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta .$$
Now let’s expand $\cos (\alpha+\beta)$ with reference to Fig. 4.12:
\begin{aligned} \cos (\alpha+\beta) &=\frac{A E}{A D}=\frac{A B-E C}{A D} \ &=\frac{A B}{A D} \frac{A C}{A C}-\frac{E C}{A D} \frac{C D}{C D} \ &=\frac{A B}{A C} \frac{A C}{A D}-\frac{E C}{C D} \frac{C D}{A D} \ \cos (\alpha+\beta) &=\cos \alpha \cos \beta-\sin \alpha \sin \beta \end{aligned}

## 计算机代写|计算机图形学代写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 \mp \tan \alpha \tan \alpha} \ \tan (2 \alpha) &=\frac{2 \tan \alpha}{1-\tan ^2 \alpha} \end{aligned}

# 计算机图形学代考

## 计算机代写|计算机图形学代写computer graphics代考|Compound-Angle Identities

$$\sin (\alpha+\beta)=\frac{F D}{A D}=\frac{B C+E D}{A D}=\frac{B C}{A D} \frac{A C}{A C}+\frac{E D}{A D} \frac{C D}{C D}=\frac{B C}{A C} \frac{A C}{A D}+\frac{E D}{C D} \frac{C D}{A D} \sin (\alpha+\beta)$$

$$\sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta .$$

$$\cos (\alpha+\beta)=\frac{A E}{A D}=\frac{A B-E C}{A D}=\frac{A B}{A D} \frac{A C}{A C}-\frac{E C}{A D} \frac{C D}{C D}=\frac{A B}{A C} \frac{A C}{A D}-\frac{E C}{C D} \frac{C D}{A D} \cos (\alpha+\beta)$$

## 计算机代写|计算机图形学代写computer graphics代考|Double-Angle Identities

$$\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \cos (\alpha \pm \beta) \quad=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \tan (\alpha \pm \beta)$$

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

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

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

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

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