Video: Evaluating Trigonometric Functions with Special Angles

In this video, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions.

18:52

Video Transcript

In this video, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions. We will begin by recalling these special angles, together with the sine, cosine, and tangent values of the angles. However, in this video, we will not consider the proofs or derivation of these as we will simply be evaluating the expressions.

When we were first introduced to the sine, cosine, and tangent trigonometric functions, we saw that they had key values at zero, 90, 180, 270, and 360 degrees. We also saw that each of the functions was periodic, as shown by their graphs. The sin of zero, 180, and 360 degrees is equal to zero. The sin of 90 degrees is equal to one. And the sin of 270 degrees is equal to negative one. This pattern continues every 90 degrees. However, for the purpose of this video, we will only be dealing with angles between zero and 360 degrees. The cos of zero degrees and 360 degrees is equal to one. The cos of 90 degrees and 270 degrees is equal to zero. And the cos of 180 degrees is equal to negative one. It is clear from the graph that the tan of zero degrees, 180 degrees, and 360 degrees is equal to zero. The tan of 90 degrees and 270 degrees are undefined, denoted by a vertical asymptote on our graph.

It is also important to note that in some of our questions, the angles will be given in radians as opposed to degrees. We know that 180 degrees is equal to 𝜋 radians. This means that 360 degrees is equal to two 𝜋 radians. 90 degrees is equal to 𝜋 over two radians. And 270 degrees is equal to three 𝜋 over two radians. As well as these key angles shown on our graphs, we also need to recall the sin, cos, and tan of 30 degrees, 45 degrees, and 60 degrees. One way of recalling these is using the table shown. In the left-hand column, we have the sine, cosine, and tangent functions. And in the top row, we have the angles 30 degrees, 45 degrees, and 60 degrees. To complete the next two rows is really straightforward. We write the numbers one, two, three and in the next row the numbers three, two, one. We then make each of these a fraction with a denominator of two.

The final step is to find the square root of the numerator. We recognize, however, that the square root of one is simply one. This gives us the value of sin 𝜃 and cos 𝜃 when 𝜃 equals 30 degrees, 45 degrees, and 60 degrees. The sin of 30 degrees is equal to one-half. The sin of 45 degrees is equal to root two over two, which is also equivalent to one over root two. The sin of 60 degrees is equal to root three over two. We also have that the cos of 30 degrees is root three over two, the cos of 45 degrees is root two over two, and the cos of 60 degrees is equal to one-half. Note at this stage that the sin of 30 degrees is equal to the cos of 60 degrees. Likewise, the cos of 30 degrees is equal to the sin of 60 degrees, and also that the sin and cos of 45 degrees are both equal to root two over two.

The values of the tan of 30, 45, and 60 degrees are not so straightforward. However, we recall the identity that the tan of angle 𝜃 is equal to the sin of angle 𝜃 divided by the cos of angle 𝜃. This means that we simply need to divide one-half by root three over two, root two over two by root two over two, and root three over two by one-half. tan of 30 degrees is therefore equal to one over root three, tan of 45 degrees is equal to one, and finally the tan of 60 degrees is equal to root three. By rationalizing the denominator, we see that one over root three is the same as root three over three. It is this value that we often use for the tan of 30 degrees.

Before looking at some examples, we will focus on the sine function. We will do this in order to spot some patterns and identify angles whose sine gives the same value. If we drew a horizontal line on the graph of 𝑦 equals sin 𝑥, where 𝑦 equals 0.5, we see that this line intersects our graph at two points between zero and 360 degrees. The equation sin 𝜃 equals 0.5 therefore has two solutions between zero and 360 degrees. We know that 0.5 is equal to one-half. And from our table, the sin of 30 degrees is equal to this.

Due to the symmetry of our graph, we see that the second angle is equal to 150 degrees. The sin of 150 degrees is also equal to 0.5 or one-half. We can use the same method to calculate the values of 𝜃 such that sin 𝜃 is equal to negative one-half. The values of 𝜃 here will be 30 degrees above 180 and 30 degrees less than 360. Therefore, the sin of 210 degrees and the sin of 330 degrees is equal to negative one-half.

An alternative way of working out these angles is using a CAST diagram. We recall that this tells us whether the sine, cosine, and tangent of an angle is positive or negative. The A stands for all, which means that all three functions have a positive value when 𝜃 lies between zero and 90 degrees. If 𝜃 lies between 90 and 180 degrees, the sine of the angle is positive, whereas the cosine and tangent is negative. When the angle lies between 180 and 270 degrees, the tangent of the angle is positive, denoted by the T, whereas the sine and cosine are negative. Finally, if 𝜃 lies between 270 and 360 degrees, the cosine of the angle is positive, whereas the sine and tangent of the angle are negative.

Using the same angle as before, we know that the sin of 30 degrees in the first quadrant is equal to one-half. In the second quadrant, the sine of the angle will also be positive. Therefore, the sin of 150 degrees also equals one-half. In the third and fourth quadrants, between 180 and 270 as well as 270 and 360 degrees, the sine of an angle is negative. This confirms that the sin of 210 degrees and the sin of 330 degrees are both equal to negative one-half. We will now use all this information to evaluate trigonometric functions where our angles are given in both radians and degrees.

Find the value of the cos of 11𝜋 over six.

We begin by noticing that our angle is given in radians, and we recall that 𝜋 radians is equal to 180 degrees. Dividing both values by six, we see that 𝜋 over six radians is equal to 30 degrees. We can then multiply both sides of this equation by 11, showing us that 11𝜋 over six radians is equal to 330 degrees. This means that the value we need to calculate is the cos of 330 degrees. We can do this by firstly sketching the graph of the cosine function and then using our knowledge of special angles.

The cosine function is periodic and has a maximum value of one and a minimum value of negative one. The graph of 𝑦 equals cos of 𝜃 between zero and 360 degrees is shown. We want to calculate the cos of 330 degrees. From our graph, we can see that this is positive and lies between zero and one. Due to the symmetry of the cosine function, we can see from our graph that the cos of 330 degrees is equal to the cos of 30 degrees. One of the special angles we need to recall is that the cos of 30 degrees is equal to root three over two. This means that the cos of 330 degrees is also equal to root three over two. The cos of 11𝜋 over six radians is, therefore, also equal to root three over two.

We will now consider a second more complicated example where the angles once again are given in radians.

Evaluate two multiplied by tan of 𝜋 by six minus eight multiplied by sin of four 𝜋 by three.

We begin by noticing that our angles here are given in radians. And we recall that 𝜋 radians is equal to 180 degrees. Dividing both of these values by six, we see that 𝜋 over six radians is equal to 30 degrees and 𝜋 over three radians is equal to 60 degrees. This means that four 𝜋 over three radians is equal to 240 degrees. Our question can, therefore, be rewritten as two multiplied by the tan of 30 degrees minus eight multiplied by the sin of 240 degrees.

Our next step is to calculate the tan of 30 degrees and the sin of 240 degrees using our knowledge of special angles and the CAST diagram. The tan of 30 degrees is equal to one over root three. This is equivalent to root three over three, and it is this form that we will use in this question. The first term in our expression is, therefore, equal to two multiplied by root three over three. Another one of our special angles gave us the result that sin of 60 degrees is equal to root three over two. We now need to consider how we can use this to calculate the sin of 240 degrees.

Sketching our CAST diagram, we see that 240 degrees lies in the third quadrant. The sine of any angle here between 180 and 270 degrees is negative. 240 degrees is equal to 180 degrees plus 60 degrees. As the horizontal or 𝑥-axis is a line of symmetry, since the sin of 60 degrees equals root three over two, the sin of 240 degrees is equal to negative root three over two. The second term of our expression becomes eight multiplied by negative root three over two. We need to subtract this from two multiplied by root three over three.

Simplifying both of our terms, our expression becomes two root three over three plus four root three. As four is equivalent to 12 over three, the second term can be rewritten as 12 root three over three. Our denominators are the same, so we can simply add the numerators, giving us 14 root three over three. Two multiplied by the tan of 30 degrees minus eight multiplied by the sin of 240 degrees is equal to 14 root three over three. This means that our original expression where the angles were given in radians is also equal to 14 root three over three.

In our final question, we will look at multiplying trigonometric functions together.

Find the value of three multiplied by the sin of 30 degrees multiplied by the sin of 60 degrees minus the cos of zero degrees multiplied by the sec of 60 degrees plus sin of 270 degrees multiplied by cos squared of 45 degrees.

In order to answer this question, we will need to recall the sine and cosine of our special angles 30, 45, and 60 degrees. We will also need to recall the reciprocal trigonometric identities, specifically the secant of any angle. Finally, we’ll need to find the product of two trigonometric functions as well as squaring the cos of 45 degrees.

The sin, cos, and tan of 30 degrees, 45 degrees, and 60 degrees can be set out in a table as shown. The sine of these three angles are equal to one-half, root two over two, and root three over two, respectively. The cos of 30, 45, and 60 degrees are equal to root three over two, root two over two, and one-half. Finally, the tangent of the three angles equals one over root three, one, and root three. We can substitute the values of sin of 30 degrees, sin of 60 degrees, and the cos of 45 degrees directly into our expression. The first term three multiplied by the sin of 30 degrees multiplied by the sin of 60 degrees is equal to three multiplied by a half multiplied by root three over two. This is equal to three root three over four.

From the graphs of 𝑦 equals sin 𝜃 and 𝑦 equals cos 𝜃, we see that the cos of zero degrees is equal to one and the sin of 270 degrees is equal to negative one. The sec of any angle 𝜃 is the reciprocal of the cos of the angle such that the sec of 𝜃 is equal to one over the cos of 𝜃. This means that the sec of 60 degrees is equal to one over the cos of 60 degrees. As the cos of 60 degrees is equal to one-half, the sec of 60 degrees is equal to two.

The second term in our expression, the cos of zero degrees multiplied by the sec of 60 degrees, is therefore equal to one multiplied by two. This is equal to two. Finally, the third term, the sin of 270 degrees multiplied by cos squared 45 degrees, is equal to negative one multiplied by root two over two squared. Root two over two squared is equal to two over four, as we simply square the numerator and denominator. This simplifies to one-half, which we need to multiply by negative one. The third term is, therefore, equal to negative one-half.

Substituting our three values into the original expression, we have three root three over four minus two plus negative one-half. Negative two plus negative one-half is the same as negative two and a half, which is equal to negative five over two. We can then multiply the numerator and denominator of our second fraction by two. This gives us three root three over four minus 10 over four. As the denominators are the same, we can subtract the numerators, giving us negative 10 plus three root three over four. This is the value of the original expression.

We will now summarize the key points from this video. We can evaluate trigonometric functions and expressions using our knowledge of special angles and the periodicity of these functions. Specifically, we can use the values of the sin, cos, and tan of 30 degrees, 45 degrees, and 60 degrees, together with the trigonometric graphs and CAST diagram.

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