Question

In Exercises 5-7, find all the exact t-values for which the given statement is true,

5. cos t= 1

6. sin t - 1/2

7. tan t = negative square root of 3

Answer 1

Answer:  See Below

Step-by-step explanation:

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

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$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does   $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees:    t = 30° + 360n  and  150° + 360n

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$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does    $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

Hint: sin and cos are only opposite signs in Quadrants II and IV

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees:    t = 120° + 360n  and  300° + 360n