Video: Simplifying Trigonometric Expressions Using Trigonometric Identities

Video Transcript

In this video, we will learn how to simplify trigonometric expressions by applying trigonometric identities. We begin by recalling that an identity is an equation that is true no matter what values are chosen. We will consider three types of trigonometric identity, the Pythagorean identities, the reciprocal identities, and the cofunction identities.

Let’s begin by considering the properties of the unit circle. We recall that the unit circle is a circle with radius one as shown. It allows us to measure the sin, cos, and tan of any angle 𝜃, as the 𝑥-coordinate of any point on the circumference represents cos 𝜃 and the 𝑦-coordinate is sin 𝜃. The right triangle in our diagram, together with the Pythagorean theorem, leads us to our first Pythagorean identity. sin squared 𝜃 plus cos squared 𝜃 equals one. Recalling the reciprocal trigonometric identities enables us to form two further Pythagorean identities.

The three reciprocal functions are the secant, cosecant, and cotangent, where the sec of angle 𝜃 is equal to one over the cos of 𝜃. The csc of 𝜃 is equal to one over sin 𝜃, and the cot of 𝜃 is equal to one over tan 𝜃. It is also worth noting that since sin 𝜃 over cos 𝜃 is equal to tan 𝜃, then cos 𝜃 over sin 𝜃 is cot 𝜃. The next Pythagorean identity is found by dividing both sides of our first identity by cos squared 𝜃. sin squared 𝜃 divided by cos squared 𝜃 is tan squared 𝜃. cos squared 𝜃 divided by cos squared 𝜃 is equal to one. And finally, one divided by cos squared 𝜃 is equal to sec squared 𝜃. Our second identity states that tan squared 𝜃 plus one is equal to sec squared 𝜃.

Returning to our first identity, we’ll now divide each term by sin squared 𝜃. This gives us one plus cot squared 𝜃 is equal to csc squared 𝜃. We now have a set of three Pythagorean identities. We will now look at a couple of examples where we need to simplify trigonometric expressions.

Simplify sin 𝜃 multiplied by csc 𝜃 minus cos squared 𝜃.

In this question, we are asked to simplify a trigonometric expression. One way of doing this is using the reciprocal and Pythagorean identities. In questions of this type, it is not always clear what we should do first. However, as a general rule, it is worth starting by replacing any reciprocal functions with the sine, cosine, and tangent functions. We know that csc 𝜃 is equal to one over sin 𝜃. Substituting this into our expression, we have sin 𝜃 multiplied by one over sin 𝜃 minus cos squared 𝜃. The sin 𝜃 on the numerator and denominator of our first term cancels. So we are left with one minus cos squared 𝜃.

Next, we recall one of our Pythagorean identities. sin squared 𝜃 plus cos squared 𝜃 is equal to one. Subtracting cos squared 𝜃 from both sides, we have sin squared 𝜃 is equal to one minus cos squared 𝜃. This means that our expression simplifies to sin squared 𝜃. sin 𝜃 multiplied by csc 𝜃 minus cos squared 𝜃 in its simplest form is sin squared 𝜃.

We will now look at a second example of this type.

Simplify sin squared 𝜃 plus cos squared 𝜃 divided by csc squared 𝜃 minus cot squared 𝜃.

In order to answer this question, we need to recall the Pythagorean identities. Firstly, we have sin squared 𝜃 plus cos squared 𝜃 is equal to one. Dividing each term by sin squared 𝜃, we have the second identity. One plus cot squared 𝜃 is equal to csc squared 𝜃. We know this based on the reciprocal trig functions where cos 𝜃 over sin 𝜃 is cot 𝜃 and one over sin 𝜃 is csc 𝜃. The numerator of our expression is the same as the left-hand side of the first identity. We can therefore replace sin squared 𝜃 plus cos squared 𝜃 with one. Subtracting cot squared 𝜃 from both sides of the second identity, we have one is equal to csc squared 𝜃 minus cot squared 𝜃.

The right-hand side here is the same as the denominator of our expression. This means that this is also equal to one. The expression simplifies to one divided by one. We can therefore conclude that sin squared 𝜃 plus cos squared 𝜃 divided by csc squared 𝜃 minus cot squared 𝜃 is equal to one.

Before looking at our next example, we will recall the cofunction identities. Once again, we consider the unit circle we saw earlier. Since the angles in a triangle sum to 180 degrees, the third angle in our right triangle is equal to 90 degrees minus 𝜃. Let’s consider what happens if we redraw this triangle such that the angle between the positive 𝑥-axis and the hypotenuse of our triangle is 90 degrees minus 𝜃. The coordinates of the point marked on the unit circle will be the cos of 90 degrees minus 𝜃, sin of 90 degrees minus 𝜃.

We notice that the distance in the 𝑥-direction here is the same as the distance in the 𝑦-direction in our first triangle. This means that the cos of 90 degrees minus 𝜃 is equal to sin 𝜃. Likewise, the sin of 90 degrees minus 𝜃 is equal to the cos of 𝜃. Since sin 𝜃 over cos 𝜃 is equal to tan 𝜃, the tangent of 90 degrees minus 𝜃 is equal to cos 𝜃 divided by sin 𝜃. And using our knowledge of the reciprocal functions, this is equal to the cot of 𝜃. It follows that the sec of 90 degrees minus 𝜃 is equal to csc 𝜃. The csc of 90 degrees minus 𝜃 is the sec of 𝜃. And the cot of 90 degrees minus 𝜃 is the tan of 𝜃.

Using our knowledge of complementary angles, we have these six equations known as the cofunction identities. It is important to note that if our angles are given in radians, the 𝜋 over two radians is equal to 90 degrees. We will now look at an example where we need to use these identities.

Simplify cos 𝜃 multiplied by csc of 90 degrees minus 𝜃 minus tan 𝜃 multiplied by tan of 90 degrees minus 𝜃.

When answering a question of this type, our first step is to try and simplify our expression using the cofunction identities. Using our knowledge of complementary angles and the unit circle, we know that the sin of 90 degrees minus 𝜃 is equal to cos 𝜃 and the cos of 90 degrees minus 𝜃 is equal to sin 𝜃. The cosecant function is the reciprocal of the sine function. It follows that the csc of 90 degrees minus 𝜃 is equal to one over cos 𝜃, which is equal to sec 𝜃. We can therefore rewrite the first term of our expression as cos 𝜃 multiplied by sec 𝜃.

Next, recalling that sin 𝜃 over cos 𝜃 is tan 𝜃, then the tan of 90 degrees minus 𝜃 is equal to cos 𝜃 over sin 𝜃, which is equal to the cot of 𝜃. The second term of our expression simplifies to tan 𝜃 multiplied by cot 𝜃.

Our next step is to use our knowledge of the reciprocal functions. We know that sec 𝜃 is equal to one over cos 𝜃 and cot 𝜃 is equal to one over tan 𝜃. Our expression becomes cos 𝜃 multiplied by one over cos 𝜃 minus tan 𝜃 multiplied by one over tan 𝜃. Both parts of our expression simplify to one. So we are left with one minus one. We can therefore conclude that cos 𝜃 multiplied by csc of 90 degrees minus 𝜃 minus tan 𝜃 multiplied by tan of 90 degrees minus 𝜃 is equal to zero.

In our final example, we will look at a more complicated problem and we’ll use a variety of the identities seen in this video.

Simplify one plus cot squared three 𝜋 over two minus 𝜃 divided by one plus tan squared three 𝜋 over two minus 𝜃.

In order to answer this question, we will need to use a variety of trigonometric identities. In problems of this type, it is not always clear where we should start. And in this question, this is complicated further by the angle three 𝜋 over two minus 𝜃. As a result, we will begin by letting 𝛼 equal three 𝜋 over two minus 𝜃. This allows us to rewrite our expression as one plus cot squared 𝛼 divided by one plus tan squared 𝛼. The Pythagorean identities state that sin squared 𝛼 plus cos squared 𝛼 is equal to one. tan squared 𝛼 plus one is equal to sec squared 𝛼. And one plus cot squared 𝛼 is equal to csc squared 𝛼.

We notice that the numerator of our fraction is the same as the left-hand side of the third identity. This means we can rewrite this as csc squared 𝛼. The denominator of our expression is the same as the left-hand side of the second identity. We can therefore rewrite the expression as csc squared 𝛼 over sin squared 𝛼. Two of the reciprocal trigonometric identities state that csc 𝛼 is equal to one over sin 𝛼 and sec 𝛼 is equal to one over cos 𝛼. The second identity could also be rewritten as one over sec 𝛼 is equal to cos 𝛼.

Rewriting csc squared 𝛼 as one over sin squared 𝛼 and one over sec squared 𝛼 as cos squared 𝛼, we have one over sin squared 𝛼 multiplied by cos squared 𝛼. This is equal to cos squared 𝛼 over sin squared 𝛼, which in turn is equal to cot squared 𝛼. Replacing 𝛼 with three 𝜋 over two minus 𝜃, we have cot squared of three 𝜋 over two minus 𝜃. We might think this is our final answer. However, we can simplify this one stage further. And this can be done by considering related angles in the unit circle. We know that the point shown on the unit circle in the first quadrant has coordinates cos 𝜃, sin 𝜃. We know that three 𝜋 over two radians is equal to 270 degrees. The point with coordinates cos of three 𝜋 over two minus 𝜃, sin of three 𝜋 over two minus 𝜃 is as shown.

We notice that the displacement in the negative 𝑥-direction of this triangle is the same as the displacement in the positive 𝑦-direction of our first triangle. This means that the cos of three 𝜋 over two minus 𝜃 is equal to the negative of sin 𝜃. Likewise, the sin of three 𝜋 over two minus 𝜃 is equal to negative cos 𝜃. Since cos 𝜃 divided by sin 𝜃 is equal to cot 𝜃, dividing these two equations we have cot of three 𝜋 over two minus 𝜃 is equal to negative sin 𝜃 over negative cos 𝜃. The right-hand side simplifies to tan 𝜃. cot squared three 𝜋 over two minus 𝜃 is therefore equal to tan squared 𝜃. The initial expression one plus cot squared three 𝜋 over two minus 𝜃 divided by one plus tan squared three 𝜋 over two minus 𝜃 is tan squared 𝜃.

We will now summarize the key points from this video. In this video, we simplified trigonometric expressions using a variety of trigonometric identities. Firstly, the Pythagorean identities. sin squared 𝜃 plus cos squared 𝜃 equals one, tan squared 𝜃 plus one equals sec squared 𝜃, and one plus cot squared 𝜃 is equal to csc squared 𝜃. We also use the reciprocal identities. csc 𝜃 is equal to one over sin 𝜃, sec 𝜃 is equal to one over cos 𝜃, and cot 𝜃 is equal to one over tan 𝜃. Since sin 𝜃 over cos 𝜃 is tan 𝜃, we also saw that cos 𝜃 over sin 𝜃 is cot 𝜃.

Finally, we saw the cofunction identities. The sin of 90 degrees minus 𝜃 is equal to cos 𝜃 and the cos of 90 degrees minus 𝜃 is sin 𝜃. Using the reciprocal identities, we’re also able to form four further cofunction identities. csc of 90 degrees minus 𝜃 is sec 𝜃, sec of 90 degrees minus 𝜃 is csc 𝜃, tan of 90 degrees minus 𝜃 is cot 𝜃, and finally cot of 90 degrees minus 𝜃 is tan 𝜃.