Question

2-77. sketch a unit circle. in your circle, sketch an angle in standard position so that the angle has: homework help a. a positive cosine and a negative sine. b. a negative cosine and a negative sine. 2-78. let f(x)=2x + 3 and h(x)=(x - 3)/2. use function composition to determine if f and h are inverse functions. how can you check your answer graphically? homework help

Explanation:

2 - 77

Step1: Recall unit - circle trigonometry

In the unit circle, for an angle $\theta$ in standard position, $\cos\theta=x$ and $\sin\theta = y$, where $(x,y)$ is the point of intersection of the terminal side of the angle and the unit circle $x^{2}+y^{2}=1$.

Step2: Find angle for positive cosine and negative sine

A positive cosine means $x>0$ and a negative sine means $y < 0$. This occurs in the fourth - quadrant. So, we can sketch an angle $\theta$ such that $270^{\circ}<\theta<360^{\circ}$ (or $\frac{3\pi}{2}<\theta < 2\pi$ in radians).

Step3: Find angle for negative cosine and negative sine

A negative cosine means $x<0$ and a negative sine means $y < 0$. This occurs in the third - quadrant. So, we can sketch an angle $\theta$ such that $180^{\circ}<\theta<270^{\circ}$ (or $\pi<\theta<\frac{3\pi}{2}$ in radians).

2 - 78

Step1: Find $f(h(x))$

Substitute $h(x)=\frac{x - 3}{2}$ into $f(x)$:
\[

$$\begin{align*} f(h(x))&=2h(x)+3\\ &=2\times\frac{x - 3}{2}+3\\ &=x-3 + 3\\ &=x \end{align*}$$

\]

Step2: Find $h(f(x))$

Substitute $f(x)=2x + 3$ into $h(x)$:
\[

$$\begin{align*} h(f(x))&=\frac{f(x)-3}{2}\\ &=\frac{(2x + 3)-3}{2}\\ &=\frac{2x}{2}\\ &=x \end{align*}$$

\]
Since $f(h(x))=x$ and $h(f(x))=x$, $f$ and $h$ are inverse functions.

Step3: Graphical check

The graphs of $y = f(x)$ and $y=h(x)$ are symmetric about the line $y = x$. The line $y = x$ has the equation $y=x$ (a straight - line with a slope of 1 and a $y$ - intercept of 0). If we plot $y = 2x+3$ (a straight - line with a slope of 2 and a $y$ - intercept of 3) and $y=\frac{x - 3}{2}=\frac{1}{2}x-\frac{3}{2}$ (a straight - line with a slope of $\frac{1}{2}$ and a $y$ - intercept of $-\frac{3}{2}$), we can see that they are symmetric about the line $y = x$.

Answer:

For 2 - 77:
a. Sketch an angle in the fourth - quadrant ($270^{\circ}<\theta<360^{\circ}$ or $\frac{3\pi}{2}<\theta < 2\pi$).
b. Sketch an angle in the third - quadrant ($180^{\circ}<\theta<270^{\circ}$ or $\pi<\theta<\frac{3\pi}{2}$).
For 2 - 78: $f$ and $h$ are inverse functions. Graphically, the graphs of $y = f(x)$ and $y = h(x)$ are symmetric about the line $y=x$.