As we've seen, the cosine and sine ratios for acute angles can be defined in the following ways:

  • Using a right triangle. If \theta is an acute angle in a right triangle, then \cos\theta = \dfrac {\text{adjacent}} {\text{hypotenuse}}, \qquad \sin\theta = \dfrac {\text{opposite}} {\text{hypotenuse}}.

  • Using the unit circle. If \theta is a central angle made by a point (x,y) on the unit circle in the first quadrant, then \cos\theta = x, \qquad \sin\theta = y.

We can't define trigonometric ratios for non-acute angles using right triangles because a right triangle can't have angles larger than 90^\circ. However, we can extend these ratios for non-acute angles using unit circles.

To do that, let's consider the unit circle with a central angle \theta in any quadrant, measured in the usual way.

We define the sine and cosine of \theta, as follows:

\cos\theta = x, \qquad \sin\theta = y

In other words,

  • the cosine of \theta is defined to be equal to the x -coordinate of P, and

  • the sine of \theta is defined to be equal to the y -coordinate of P.

This way, our trigonometric ratios make sense for any angle, even if this angle is larger than 90^\circ.

FLAG

The diagram above shows a unit circle. Given that the x -coordinate of the point P is -0.7, what is the value of \cos\theta?

EXPLANATION

Any point (x,y) on the unit circle is related to the central angle \theta as follows:

x = \cos\theta, \qquad y = \sin\theta

We're given that x=-0.7 at the point P. Therefore, we have

\cos\theta = -0.7.

FLAG

The diagram above shows a unit circle. Given that the $y$-coordinate of the point $P$ is $-0.75,$ which of the following is correct?

a
$\sin\theta = 0.15$
b
$\cos\theta = 0.15$
c
$\cos\theta = -0.75$
d
$\sin\theta = 0.75$
e
$\sin\theta = -0.75$

The diagram above shows a unit circle. Given that the $x$-coordinate of the point $P$ is $0.55,$ which of the following is correct?

a
$\cos\theta = -0.55$
b
$\cos\theta = 0.55$
c
$\sin\theta = 0.55$
d
$\sin\theta = 0.45$
e
$\cos\theta = 0.45$

We can use reference angles to evaluate trigonometric ratios for angles outside the first quadrant.

To demonstrate, let's determine the value of \cos (135^\circ), corresponding to the x -coordinate of the point P on the unit circle, as shown below.

The reference angle for \theta=135^\circ is \theta_R=45^\circ. Let's add this to our diagram.

Let P' be the image of P under a reflection across the y -axis. This gives a pair of congruent angles of measure 45^\circ, as shown.

According to the definition, \cos(135^\circ) is the x -coordinate of the point P. But this is the same as the x -coordinate of the point P' taken with the opposite sign.

The x -coordinate of P' equals the cosine of the reference angle. \cos(45^\circ)=\dfrac{\sqrt{2}}{2} Therefore, we conclude that \cos(\, \underbrace{135^\circ}_{\theta} \,) = - \cos(\, \underbrace{45^\circ}_{\theta_R} \,) = -\dfrac{\sqrt{2}}{2}.

So, we simplified the evaluation of \cos(135^\circ) by evaluating the cosine of the corresponding reference angle \theta_R=45^\circ with an additional sign adjustment for the final result.

We can use a similar argument to show that, since y = \sin\theta, we have \sin(135^\circ) = \sin(45^\circ) = \dfrac{\sqrt 2}{2}.

In this case, we do NOT change the sign because the y -coordinates of P' and P are equal.

FLAG

We can calculate a trigonometric ratio of any angle \theta by finding the ratio of the corresponding reference angle and adjusting the sign depending on which quadrant our angle lies in.

To do this, we perform the following steps:

  1. Find the reference angle \theta_R.

  2. Calculate the value of the ratio for \theta_R.

  3. Determine whether the resulting ratio is positive or negative.

We must do the last step because the ratio for \theta_R will always be nonnegative, but the ratio for \theta may not be.

But how do we determine whether the ratio for \theta is positive or negative? This depends on the trigonometric ratio and which quadrant the angle is in.

To help us remember when various trigonometric ratios are positive, we use the mnemonic "All Students Take Calculus."

  • All of the trigonometric ratios are positive in quadrant I

  • Sine (and only sine) is positive in quadrant II

  • Tangent (and only tangent) is positive in quadrant III

  • Cosine (and only cosine) is positive in quadrant IV

We can represent this in a so-called CAST diagram, shown below.

For example, let's consider \sin 140^\circ.

Step 1: Since \theta = 140^\circ is the 2 nd quadrant, the reference angle \theta_R is

\begin{align*} \theta_R &= 180^\circ - \theta\\[5pt] &= 180^\circ - 140^\circ\\[5pt] &= 40^\circ. \end{align*}

Step 2: The given ratio is \sin{140^\circ}, and therefore we're interested in \sin\theta_R = \sin{40^\circ}.

Step 3: The ratio \sin{140^\circ} must be positive because the sine ratio is always positive in the 2 nd quadrant. Therefore,

\sin 140^\circ = \sin{40^\circ}.

Let's see another example.

FLAG

What is \sin 335^\circ expressed in terms of a reference angle?

EXPLANATION

To express \sin{335^\circ} in terms of \sin\theta_R, we follow three steps:

  1. Find its reference angle \theta_R.

  2. Calculate the value of the function for \theta_R.

  3. Determine whether the resulting value is positive or negative.

First, let's draw the angle 335^\circ in the coordinate plane (CAST diagram):

Step 1: Since \theta = 335^\circ is in the 4 th quadrant, the reference angle \theta_R is

\begin{align*} \theta_R &= 360^\circ - \theta\\[3pt] &= 360^\circ - 335^\circ\\[3pt] &= 25^\circ. \end{align*}

Step 2: The given ratio is \sin{335^\circ}, and therefore we're interested in \sin\theta_R = \sin{25^\circ}.

Step 3: The ratio \sin{335^\circ} must be negative because the sine ratio is always negative in the 4 th quadrant. Therefore,

\sin 335^\circ = -\sin{25^\circ}.

FLAG

What is $\sin 175^\circ$ expressed in terms of a reference angle?

a
$\sin 185^\circ$
b
$-\sin 5^\circ$
c
$\cos 185^\circ$
d
$\sin 355^\circ$
e
$\sin 5^\circ$

What is $\sin 320^\circ$ expressed in terms of a reference angle?

a
$-\sin 320^\circ$
b
$-\cos 320^\circ$
c
$-\sin 40^\circ$
d
$\sin 40^\circ$
e
$\cos 50^\circ$

What is \cos 340^\circ expressed in terms of a reference angle?

EXPLANATION

To express \cos{340^\circ} in terms of \cos\theta_R, we follow three steps:

  1. Find its reference angle \theta_R.

  2. Calculate the value of the function for \theta_R.

  3. Determine whether the resulting value is positive or negative.

First, let's draw the angle 340^\circ in the coordinate plane (CAST diagram):

Step 1: Since \theta = 340^\circ is in the 4 th quadrant, the reference angle \theta_R is

\begin{align*} \theta_R &= 360^\circ - \theta\\ &= 360^\circ - 340^\circ\\ &= 20^\circ. \end{align*}

Step 2: The given ratio is \cos{340^\circ}, and therefore we're interested in \cos\theta_R = \cos{20^\circ}.

Step 3: The ratio \cos{340^\circ} must be positive because the cosine ratio is always positive in the 4 th quadrant. Therefore,

\cos 340^\circ = \cos{20^\circ}.

FLAG

What is $\cos 300^\circ$ expressed in terms of a reference angle?

a
$-\cos 60^\circ$
b
$-\sin 60^\circ$
c
$\cos 300^\circ$
d
$\cos 60^\circ$
e
$\sin 60^\circ$

What is $\cos 250^\circ$ expressed in terms of a reference angle?

a
$\cos 250^\circ$
b
$\cos 70^\circ$
c
$\sin 70^\circ$
d
$-\sin 70^\circ$
e
$-\cos 70^\circ$

What is \tan 162^\circ expressed in terms of a reference angle?

EXPLANATION

To express \tan{162^\circ} in terms of \tan\theta_R, we follow three steps:

  1. Find its reference angle \theta_R.

  2. Calculate the value of the function for \theta_R.

  3. Determine whether the resulting value is positive or negative.

First, let's draw the angle 162^\circ in the coordinate plane (CAST diagram):

Step 1: Since \theta = 162^\circ is in the 2 nd quadrant, the reference angle \theta_R is

\begin{align*} \theta_R &= 180^\circ - \theta\\[3pt] &= 180^\circ-162^\circ\\[3pt] &= 18^\circ. \end{align*}

Step 2: The given ratio is \tan{162^\circ} , and therefore we're interested in \tan\theta_R = \tan{18^\circ}.

Step 3: The ratio \tan{162^\circ} must be negative because the tangent ratio is always negative in the 2 nd quadrant. Therefore,

\tan 162^\circ = -\tan{18^\circ}.

FLAG

What is $\tan 235^\circ$ expressed in terms of a reference angle?

a
$\tan{55^\circ}$
b
$\tan{35^\circ}$
c
$-\tan{55^\circ}$
d
$-\tan{235^\circ}$
e
$-\tan{125^\circ}$

What is $\tan 325^\circ$ expressed in terms of a reference angle?

a
$-\tan{55^\circ}$
b
$-\tan{325^\circ}$
c
$\tan{35^\circ}$
d
$-\tan{35^\circ}$
e
$\tan{55^\circ}$
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