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**Section 2.3 #35:** The values of $\sin(\theta)$ and $\cos(\theta)$ are given. Find the value of the four remaining trigonometric functions: $\sin(\theta)=-\dfrac{3}{5}$ and $\cos(\theta)=\dfrac{4}{5}$.

*Solution:* Calculate
$$\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{-\frac{3}{5}}{\frac{4}{5}}=\left( -\dfrac{3}{5} \right) \left( \dfrac{5}{4} \right) = -\dfrac{3}{4} ,$$
$$\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{\cos(\theta)}{\sin(\theta)} = \dfrac{\frac{4}{5}}{-\frac{3}{5}} = \left( \dfrac{4}{5} \right) \left( - \dfrac{5}{3} \right) = -\dfrac{4}{3}, $$
$$\csc(\theta) = \dfrac{1}{\sin(\theta)} = \dfrac{1}{-\frac{3}{5}} = -\dfrac{5}{3},$$
and
$$\sec(\theta) = \dfrac{1}{\cos(\theta)} = \dfrac{1}{\frac{4}{5}} = \dfrac{5}{4}.$$

**Section 2.3 #44:** Find the value of the remaining five trigonometric functions given that $\cos(\theta) = \dfrac{3}{5}$ and $\theta$ is in quadrant IV.

*Solution:* We will first find the value of $\sin(\theta)$ using the Pythagorean identity for $\sin$ and $\cos$, $\sin^2(\theta)+\cos^2(\theta)=1$. Plugging into this the known value $\cos(\theta)=\dfrac{3}{5}$ yields the equation
$$\sin^2(\theta) + \left(\dfrac{3}{5} \right)^2 = 1.$$
Isolate $\sin^2(\theta)$ to get
$$\sin^2(\theta) = 1 - \dfrac{9}{25} = \dfrac{16}{25}.$$
Take the square root of each side of the equation (recall when doing this you must consider both the positive *and* negative square root):
$$\sin(\theta) = \pm \sqrt{\dfrac{16}{25}} = \pm \dfrac{4}{5}.$$
But we are told in the statement of the problem that $\theta$ is in quadrant IV and so $\sin(\theta)$ must be negative. Hence we take the value
$$\sin(\theta) = -\dfrac{4}{5}.$$
From this we may compute
$$\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}=\dfrac{-\frac{4}{5}}{\frac{3}{5}}=\left( - \dfrac{4}{5} \right) \left( \dfrac{5}{3} \right) = -\dfrac{4}{3},$$
$$\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{\cos(\theta)}{\sin(\theta)} =\dfrac{\frac{3}{5}}{-\frac{4}{5}} = \left( \dfrac{3}{5} \right) \left( - \dfrac{5}{4} \right) = -\dfrac{3}{4},$$
$$\csc(\theta) = \dfrac{1}{\sin(\theta)} = \dfrac{1}{-\frac{4}{5}} = -\dfrac{5}{4},$$
and
$$\sec(\theta) = \dfrac{1}{\cos(\theta)} = \dfrac{1}{\frac{3}{5}} = \dfrac{5}{3}.$$

**Section 2.3 #56:** Find the value of the remaining five trigonometric functions given that $\cot(\theta) = \dfrac{4}{3}$ and $\cos(\theta)<0$.

*Solution:* We use the Pythagorean identity for cotangent and cosecant: $\cot^2(\theta) + 1 = \csc^2(\theta)$. Substitue the value we know for $\cot(\theta)$ to get the equation
$$\left( \dfrac{4}{3} \right)^2 + 1 = \csc^2(\theta).$$
Simplify the left-hand side to get
$$\dfrac{25}{9} = \csc^2(\theta),$$
so take the square root of both sides (be careful to include the $\pm$!) to get
$$\csc(\theta) = \pm \dfrac{5}{3}.$$
How do we determine which to take? We were told that $\cos(\theta)<0$ and so $\theta$ must lie in quadrants II or III. But we also know that $\cot(\theta)=\dfrac{4}{3}$, and so since cotangent is positive only in quadrants I and III, we must deduce that $\theta$ lies in quadrant III. Since sine is negative in quadrant III and $\csc(\theta) = \dfrac{1}{\sin(\theta)}$, we must take the value of $\csc(\theta)$ to be $-\dfrac{5}{3}$. Therefore
$$\sin(\theta) = \dfrac{1}{\csc(\theta)} = -\dfrac{3}{5},$$
$$\tan(\theta) = \dfrac{1}{\cot(\theta)} = \dfrac{3}{4}.$$
To find $\cos(\theta)$ and $\sec(\theta)$ note that we know $\sin(\theta)$ and $\cos(\theta)$ and since $\cot(\theta) =\dfrac{\cos(\theta)}{\sin(\theta)}$ we may multiply both sides of that equation by $\sin(\theta)$ to get the formula $\cot(\theta) \sin(\theta) = \cos(\theta)$. Therefore we see
$$\cos(\theta) = \cot(\theta) \sin(\theta) = \dfrac{4}{3} \cdot \left( - \dfrac{3}{5} \right) = -\dfrac{12}{15}$$
and
$$\sec(\theta) = \dfrac{1}{\cos(\theta)} = -\dfrac{15}{12}.$$

**Section 2.3 #72:** Use the even-odd properties to find the exact value of $\sin \left( -\dfrac{3\pi}{2} \right)$.

*Solution:* Recall that sine is an odd function and so $\sin(-x)=-\sin(x)$. We know from the unit circle that $\sin \left( \dfrac{3\pi}{2} \right)=-1$. Therefore we may compute
$$\sin \left( - \dfrac{3\pi}{2} \right) = -\sin \left( \dfrac{3\pi}{2} \right) = -(-1) = 1.$$

**Section 2.3 #113:** If $f(\theta)=\sin(\theta)$ and $f(a)=\dfrac{1}{3}$, find the exact value of

a.) $f(-a)$, and

b.) $f(a)+f(a+2\pi)+f(a+4\pi)$.

*Solution:* For part a.), since sine is an odd function, we may compute
$$f(-a)=-f(a)=-\dfrac{1}{3}.$$
For part b.), since sine is periodic with period $2\pi$, we may compute
$$f(a)+f(a+2\pi)+f(a+4\pi) = f(a)+f(a)+f(a+2\pi) = 2f(a)+f(a) = 3f(a) = 1.$$