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解析几何和三角学|Analytic Geometry and Trigonometry代写 MATH 102

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这是一份umass麻省大学 MATH 102作业代写的成功案例

解析几何和三角学|Analytic Geometry and Trigonometry代写 MATH 102
问题 1.

Similarly,
$$
\cos \left(180^{\circ}-\theta\right)=\frac{x^{\prime}}{r^{\prime}}=\frac{-x}{r}=-\frac{x}{r}=-\cos \theta .
$$
$$
\therefore \cos \left(180^{\circ}-\theta\right)=-\cos \theta \text {. }
$$
Similarly,
$$
\begin{gathered}
\tan \left(180^{\circ}-\theta\right)=\frac{y^{\prime}}{x^{\prime}}=\frac{y}{-x}=-\frac{y}{x}=-\tan \theta . \
\therefore \tan \left(180^{\circ}-\theta\right)=-\tan \theta .
\end{gathered}
$$

证明 .

In like fashion,
$\csc \left(180^{\circ}-\theta\right)=\csc \theta$ $\sec \left(180^{\circ}-\theta\right)=-\sec \theta$ $\cot \left(180^{\circ}-\theta\right)=-\cot \theta$ Example
Find: $\sin 150^{\circ} ; \cot 135^{\circ}$.
$$
\begin{aligned}
&\sin 150^{\circ}=\sin \left(180^{\circ}-30^{\circ}\right)=\sin 30^{\circ}=\frac{1}{2} . \
&\cot 135^{\circ}=\cot \left(180^{\circ}-45^{\circ}\right)=-\cot 45^{\circ}=-1
\end{aligned}
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH102 COURSE NOTES ：

We next solve triangle $A B^{\prime} C$, in Figure 32 . Let $A B^{\prime}=c^{\prime}$, and $\angle A C B^{\prime}=C^{\prime}$.
$$
\begin{gathered}
B=57^{\circ} 1^{\prime} \
\therefore B^{\prime}=180^{\circ}-57^{\circ} 1^{\prime}=122^{\circ} 59^{\prime} \
\therefore C^{\prime}=\angle A C B^{\prime}=180^{\circ}-\left(122^{\circ} 59^{\prime}+34^{\circ}\right)=23^{\circ} 1^{\prime}
\end{gathered}
$$




预微积分、代数、函数和图形|Precalculus, Algebra, Functions and Graphs代写 MATH 101

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这是一份umass麻省大学 MATH 101作业代写的成功案例

预微积分、代数、函数和图形|Precalculus, Algebra, Functions and Graphs代写 MATH 101
问题 1.

$$
\beta: A \times B \longrightarrow \operatorname{Hom}{T}\left(C, A \otimes{R}\left(B \otimes_{S} C\right)\right)
$$
is a bihomomorphism of bimodules, and there is a unique homomorphism
$$
\bar{\beta}: A \otimes_{R} B \longrightarrow \operatorname{Hom}{T}\left(C, A \otimes{R}\left(B \otimes_{S} C\right)\right)
$$

证明 .

of bimodules such that $\bar{\beta}(a \otimes b)=\beta(a, b)$ for all $a, b$. By $5.6,(u, c) \longmapsto \bar{\beta}(u)(c)$ is a bihomomorphism of $\left(A \otimes_{R} B\right) \times C$ into $A \otimes_{R}\left(B \otimes_{S} C\right)$. Hence there is a bimodule homomorphism
$$
\theta:\left(A \otimes_{R} B\right) \otimes_{S} C \longrightarrow A \otimes_{R}\left(B \otimes_{S} C\right)
$$

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MATH101 COURSE NOTES ：

Proof. The set Bihom $(A \times B, C)$ of all bihomomorphisms of $A \times B$ into $C$ is an abelian group under pointwise addition. The universal property of the tensor map $\tau:(a, b) \longmapsto a \otimes b$ provides a bijection $\varphi \longmapsto \varphi \circ \tau \circ \operatorname{Hom}{\mathbb{Z}}\left(A \otimes{R} B, C\right)$ onto Bihom $(A \times B, C)$, which preserves pointwise addition. Proposition $5.2$ provides two more bijections:
$\operatorname{Bihom}(A \times B, C) \longrightarrow \operatorname{Hom}{R}\left(A, \operatorname{Hom}{Z}(B, C)\right)$,
$\operatorname{Bihom}(A \times B, C) \longrightarrow \operatorname{Hom}{R}\left(B, \operatorname{Hom}{\mathcal{Z}}(A, C)\right)$,