# Lesson Video: Simplifying Trigonometric Expressions Mathematics

In this video, we will learn how to simplify a trigonometric expression.

16:21

### Video Transcript

In this video, we will learn how to simplify trigonometric expressions. We will recap the reciprocal trigonometric functions and recall which functions are odd and which are even. We’ll then look at the Pythagorean identities and the cofunction identities and learn how to use these to simplify expressions. Let’s start by recapping some reciprocal trigonometric functions that we should already know.

csc of 𝜃 is equal to one over sin of 𝜃, and sec of 𝜃 is equal to one over the cos of 𝜃. We can of course see any of these functions or identities written with the variable of 𝑥 instead of 𝜃, but the same rule applies. We also have the function that the tan of 𝜃 is equal to sin of 𝜃 over cos of 𝜃 and the cot of 𝜃 is equal to one over the tan of 𝜃. Because the tan of 𝜃 is equal to sin of 𝜃 over cos of 𝜃, then the cot of 𝜃 can also be written as the cos of 𝜃 divided by the sin of 𝜃.

We can also recall the identities for negative angles. Because sine and tangent are odd functions, then the sin of negative 𝜃 is equal to negative sin 𝜃 and tan of negative 𝜃 is equal to negative tan of 𝜃. Cosine is an even function, so the cos of negative 𝜃 is equal to the cos of 𝜃.

We’ll now look at a few questions where we use these trigonometric identities to simplify expressions.

Find the value of eight over sin 𝜃 multiplied by negative five over csc 𝜃.

When we have an expression like this involving a cosecant, then it’s always useful to remember the reciprocal identity. csc 𝜃 is equal to one over sin 𝜃. So when we simplify this expression, we’ll replace csc 𝜃 with one over sin 𝜃. When it comes to simplifying this fraction, negative five over one over sin 𝜃, it’s worthwhile remembering that this is equivalent to negative five divided by one over sin 𝜃. And to divide by a fraction, we multiply by its reciprocal. So we’ll be working out negative five multiplied by sin 𝜃 over one, which gives us negative five sin 𝜃.

So now when we return to the original calculation, we’ll be working out eight over sin 𝜃 multiplied by negative five sin 𝜃. Considering the second value as a fraction over one, we then realize that we can divide through by sin of 𝜃. And so we’re left with eight multiplied by negative five divided by one, which gives us negative 40 over one or just simply negative 40. Therefore, we’ve answered the question. The value of eight over sin 𝜃 multiplied by negative five over csc 𝜃 is equal to negative 40.

Let’s take a look at another question.

Simplify cos squared 𝜃 sec 𝜃 csc 𝜃.

When we have questions like this that involve a large number of functions, it can initially be quite confusing. However, the best thing to do is think about the different functions that make up our expression and see if we can recall any identities regarding these. As we have sec of 𝜃 and csc of 𝜃, it might be useful to recall that the sec of 𝜃 is equal to one over the cos of 𝜃 and the csc of 𝜃 is equal to one over sin of 𝜃. We can then replace these two functions with their reciprocal functions, which gives us cos squared 𝜃 multiplied by one over cos of 𝜃 multiplied by one over sin 𝜃.

When we simplify this expression, we have cos squared 𝜃 over cos 𝜃 sin 𝜃. At this point, it might be worth reminding ourselves that cos squared 𝜃 is just the same as cos 𝜃 multiplied by cos 𝜃. Now, when we look at this expression, we should notice that we can take out a common factor of cos 𝜃 from the numerator and denominator. This leaves us with cos 𝜃 over sin 𝜃, and this expression should look familiar. The cot of 𝜃 is in fact equal to cos of 𝜃 over sin of 𝜃. And so what we have here is the cot of 𝜃. And so our answer is that cos squared 𝜃 sec 𝜃 csc 𝜃 simplifies to cot 𝜃.

In the next question, we’ll need to use some of the even and odd identities to help us simplify a trigonometric expression.

Simplify tan of negative 𝜃 csc of 𝜃.

When we’re simplifying trigonometric expressions, if we have a look at the question, it will give us a clue as to which identity we’ll need to use. The angle of negative 𝜃 here should make it clear that we probably need to use the even and odd identities or often called the negative angle identities. The tangent function is an odd function. So that means that tan of negative 𝜃 is equal to negative tan of 𝜃. As this function also involves the csc of 𝜃, it’s worthwhile remembering the reciprocal identity. The csc of 𝜃 is equal to one over sin of 𝜃.

We can then plug in these values. The tan of negative 𝜃 is equal to negative tan of 𝜃, and csc of 𝜃 is equal to one over sin of 𝜃. Sometimes when we’re simplifying trigonometric expressions, the best thing we can do is to see if we can write our expressions in terms of sine or cosine. If we look at this negative tan of 𝜃, we can in fact replace this with something else. The tan of 𝜃 is equal to sin 𝜃 over cos 𝜃.

We can then plug this value into our expression for the tan of 𝜃, being careful that we don’t lose this negative sign along the way. Before we simplify negative sin 𝜃 over cos of 𝜃 multiplied by one over sin 𝜃, observe that we can take out a common factor of sin of 𝜃 from the numerator and denominator. Multiplying what remains then, we have negative one over cos of 𝜃.

We should remember that sec of 𝜃 is equal to one over cos of 𝜃. But as we have negative one over cos of 𝜃, the answer will be negative sec of 𝜃. And so we’ve simplified tan of negative 𝜃 csc of 𝜃 to negative sec of 𝜃.

Before we look at any more questions, let’s have a look at the Pythagorean identity for trigonometry. It’s perhaps the most commonly used of all the trigonometric identities. And it tells us that sin squared 𝜃 plus cos squared 𝜃 is equal to one. We can of course rearrange this into different formats. For example, by subtracting cos squared 𝜃 from both sides, we would find that sin squared of 𝜃 is equal to one minus cos squared of 𝜃. We can even manipulate it by dividing through by cos squared 𝜃.

Simplifying the first term of sin squared 𝜃 over cos squared 𝜃 gives us tan squared 𝜃. The next term of cos squared 𝜃 over cos squared 𝜃 would simplify to one. On the right-hand side, one over cos squared 𝜃 simplifies to sec squared 𝜃. And we know this because one over cos 𝜃 is equal to sec 𝜃. Knowing both of these identities off by heart will allow us to very quickly manipulate and simplify trigonometric expressions.

We’ll now have a look at some questions where we’ll use these identities along with the other ones that we already know.

Simplify sin 𝜃 plus cos 𝜃 squared minus two sin 𝜃 cos 𝜃.

When we first look at this question, it might seem a little more complicated because we don’t know which identities we’ll need to use. However, let’s see if we can do anything with this sin 𝜃 plus cos 𝜃 squared. When we square a value, it just means we’re multiplying it by itself. We can then expand these parentheses using a method like the FOIL method. sin of 𝜃 multiplied by sin of 𝜃 gives us sin squared 𝜃. We have sin of 𝜃 multiplied by cos of 𝜃 two times. And finally cos 𝜃 multiplied by cos 𝜃 gives us cos squared 𝜃.

An alternative to multiplying out the double brackets in this way is to remember that if we have this bracket multiplied, we square the first term, we add on the square of the second term, and then we add on double the product of both terms. Either method would give us sin squared 𝜃 plus two sin 𝜃 cos 𝜃 plus cos squared 𝜃 minus two sin 𝜃 cos 𝜃. When we look at this expression, we can observe we have two sin 𝜃 cos 𝜃 minus two sin 𝜃 cos 𝜃, which is equivalent to zero.

Now, all that we’re left with is sin squared 𝜃 plus cos squared 𝜃. And this expression should be familiar. The Pythagorean identity tells us that sin squared 𝜃 plus cos squared 𝜃 is equal to one. This is exactly the same as we’ve found in our workings. And so we can give the answer that sin 𝜃 plus cos 𝜃 squared minus two sin 𝜃 cos 𝜃 simplifies to one.

Let’s have a look at another question involving this Pythagorean identity.

Simplify sin 𝜃 csc 𝜃 minus cos squared 𝜃.

When we look at a question like this, perhaps the most obvious place to start would be to see if we can rewrite csc of 𝜃 in a different way. We can remember that the csc of 𝜃 is equal to one over sin 𝜃. When we replace csc of 𝜃 with one over sin 𝜃, we end up with the expression sin of 𝜃 multiplied by one over sin of 𝜃 minus cos squared 𝜃. Working out sin 𝜃 multiplied by one over sin 𝜃 will give us sin of 𝜃 over sin of 𝜃. And this of course simplifies to one. So how do we simplify one minus cos squared 𝜃 any further?

Well, we can remember the Pythagorean identity, which tells us that sin squared 𝜃 plus cos squared 𝜃 is equal to one. Rearranging this equation by subtracting cos squared 𝜃 from both sides would give us that sin squared 𝜃 is equal to one minus cos squared 𝜃. And this is exactly what we have in our simplified expression. One minus cos squared 𝜃 is equal to sin squared 𝜃. And so we’ve got the answer that sin 𝜃 csc 𝜃 minus cos squared 𝜃 can be simplified to sin squared 𝜃.

We’ll now look at a different set of trigonometric identities, the cofunction identities, which is when we express trigonometric functions in terms of their compliments. So we have cos of 𝜃 equals sin of 𝜋 over two minus 𝜃, sin of 𝜃 equals cos of 𝜋 over two minus 𝜃, cot of 𝜃 equals tan of 𝜋 over two minus 𝜃, tan of 𝜃 equals cot of 𝜋 over two minus 𝜃, the csc of 𝜃 equals sec of 𝜋 over two minus 𝜃, and finally sec 𝜃 equals csc of 𝜋 over two minus 𝜃.

It’s worth pointing out here that all of these angles will be expressed in radians. But we could also express them in terms of degrees. For example, the cos of 𝜃 is equal to sin of 90 degrees minus 𝜃. It may be worth pausing the video to make a note of these identities as we’ll be using some of them in the final question.

Simplify sec of 𝜋 over two minus 𝜃 over cot of 𝜋 minus 𝜃.

When we have a trigonometric expression which involves an angle of 𝜋 over two minus 𝜃, it’s a good indication that we’ll probably be using the cofunction identities. The cofunction identity for sec of 𝜋 over two minus 𝜃 tells us that this is equivalent to csc of 𝜃. Therefore, when we simplify this expression, the numerator will be csc of 𝜃. So what about this denominator?

Well, one possible way to simplify this is to remember that the cot of 𝜃 is equal to cos 𝜃 over sin of 𝜃. So for our value of cot 𝜋 minus 𝜃, this will become cos of 𝜋 minus 𝜃 over sin of 𝜋 minus 𝜃.

Next, let’s use the fact that cos of 𝜋 minus 𝜃 equals negative cos 𝜃 and sin of 𝜋 minus 𝜃 equals sin 𝜃. Therefore, our expression has csc 𝜃 on the numerator and negative cos 𝜃 over sin 𝜃 on the denominator. Before we tidy up this fraction on the denominator, let’s see if we can do anything more with the csc of 𝜃 on the numerator.

We should remember one of the reciprocal identities, which tells us that the csc of 𝜃 is equal to one over sin of 𝜃. Therefore, on the numerator, we can replace csc of 𝜃 with one over sin 𝜃. This expression now does look very confusing with a fraction on the numerator and denominator. It can be helpful to remember that a fraction simply means the numerator divided by the denominator.

Remember that when we’re dividing fractions, we simply multiply by the reciprocal of the second fraction. We notice that before we multiply, we can take out this common factor of sin 𝜃 from the numerator and denominator. Multiplying these values then gives us one over negative cos 𝜃. We can use one final trigonometric identity. The sec of 𝜃 is equal to one over the cos of 𝜃. As we have one over negative cos 𝜃, then our value will simplify to negative sec 𝜃. And so we’ve answered the question to simplify sec of 𝜋 over two minus 𝜃 over cot of 𝜋 minus 𝜃.

We can now recap some of the identities that we saw in this video. We began this video by recapping some of the reciprocal identities. We then reminded ourselves of the negative angle or odd and even function identities. We looked at the Pythagorean identity sin squared 𝜃 plus cos squared 𝜃 equals one. And we also saw how we can use this to derive sec squared 𝜃 equals one plus tan squared 𝜃. Finally, we also saw the cofunction identities. And finally, it’s worth remembering that we often need to apply more than one identity or even type of identity to simplify a trigonometric expression.

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