Proving Trig Identities
Proving Trig Identities – Introduction
Proving trig identities is an important topic in AS Mathematics. It involves demonstrating that an equation involving trigonometric functions holds true for all values of the variables involved. The use of common trigonometric identities is crucial in this process.
Common identities such as the Pythagorean identity \sin ^2 \theta+\cos ^2 \theta \equiv 1 and the reciprocal identities (\operatorname{cosec} \theta=1 / \sin \theta, \sec \theta=1 / \cos \theta, \text { and } \cot \theta=1 / \tan \theta) are frequently employed when proving trigonometric identities at AS level. These identities help simplify expressions and establish connections between different trigonometric functions.
It is worth noting that proving trig identities allows for multiple approaches, and there is no single right or wrong way to do it. Students can choose from various techniques, such as manipulating one side of the equation to match the other side or simplifying both sides using trigonometric identities until they are equivalent.
In A2 level, which corresponds to year 13, the complexity of trigonometric identities increases. Students are introduced to more advanced identities, including the sum and difference formulas, double angle formulas, and product-to-sum identities. These identities expand the range of equations that can be proven and provide a deeper understanding of trigonometry.
Overall, proving trigonometric identities is an essential skill in AS Mathematics, requiring the use of common identities and allowing for various approaches. It further develops in A2 level, enabling students to tackle more complex equations and delve deeper into the realm of trigonometry.
One of the biggest topics that you will study at A Level is that of Trigonometry. You will have met trigonometry at GCSE and whilst you will still need to know about the Sine and Cosine Rule as well as Pythagoras’s Theorem you will also meet a new area of trigonometry.
This area is that of proving trig identities and you will see more in year 13 with the use of reciprocal trigonometric functions compared to that seen in year 12.
This article is going to consider proving trig identities for year 12 only. There will be another article that involves the year 13 trig identities at a later date.
In order to prove trig identities there are two trigonometric identities that you need to remember and know when to use. These are:
\begin{aligned} & \tan \theta \equiv \frac{\sin \theta}{\cos \theta} \\ & \sin ^2 \theta+\cos ^2 \theta \equiv 1 \end{aligned}Simple Proving Trig Identities - Examples
Part a)
We want to simply \sin ^2 3 \theta+\cos ^2 3 \theta
Because \sin ^2 \theta+\cos ^2 \theta=1 then \theta \text { replaced by } 3 \theta \text {. } and so the final answer is just 1.
Part b)
We want to simplify 5-5 \sin ^2 \theta
As well as using trig identities it is also important to look out for factors. In this a common factor is 5 and so the above can be factorised. Once factorised you can then use a trig identity to then fully simplify the expression which is shown below:
\begin{gathered} =5\left(1-\sin ^2 \theta\right) \\ =5 \cos ^2 \theta . \end{gathered}Part c)
We want to simplify this slightly more complicated expression
\frac{\sin 2 \theta}{\sqrt{1-\sin ^2 2 \theta}}In terms of where to start it, in this case it is the denominator and we are using the following fact:
\sin ^2 2 \theta+\cos ^2 2 \theta=1 \text {, so } 1-\sin ^2 2 \theta=\cos ^2 2 \theta \text {. }Having obtained this result we can then continue with the full simplification as follows:
\begin{aligned} & =\frac{\sin 2 \theta}{\sqrt{\cos ^2 2 \theta}} \\ & =\frac{\sin 2 \theta}{\cos 2 \theta} \\ & =\tan 2 \theta \end{aligned}The above question has relied upon using both trig identities and it is quite common to have to use both.
Proving Trig Identities - Example
Show that (\sin \theta+\cos \theta)(\sin \theta-\cos \theta) \equiv 2 \sin ^2 \theta-1 \text {. }
Given that the question has said “show that” your aim is to start with either the left side or the right side and to show that it is equal to the other side. You need to understand that you are not solving an equation.
Now the issue with trig identity questions is that there is no right or wrong way to do the questions. Sometimes you can start on the right and prove that it is the same as the left and there can be instances where you start on the right and come to a halt. The same also applies if you start on the left.
Unfortunately you are not going to know which side is best until you actually start doing the question. Sometimes it can be a case of just trying a different approach to the question but it is a case of “trying something” and seeing what happens.
Going back to our original question you can see that on the left there are brackets. It would be best to expand these rather than starting with the right side because you just have more to work with.
So starting with the left side and expanding the brackets will give:
(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)=\sin ^2 \theta-\cos ^2 \thetaNow in order to proceed it is a case of using a trig identity and this is as follows:
\sin ^2 \theta+\cos ^2 \theta \equiv 1 \text { then } \cos ^2 \theta \equiv 1-\sin ^2 \thetaYou can then perform a substitution and this is done as follows:
\begin{aligned} & (\sin \theta+\cos \theta)(\sin \theta-\cos \theta) \equiv \sin ^2 \theta-\left(1-\sin ^2 \theta\right) \\ & (\sin \theta+\cos \theta)(\sin \theta-\cos \theta) \equiv 2 \sin ^2 \theta-1 \end{aligned}
Proving Trig Identities – Harder Example
Prove that \frac{\cos ^4 \theta-\sin ^4 \theta}{\cos ^2 \theta} \equiv 1-\tan ^2 \theta
With questions like this the most common question that is asked is knowing with which side to start. Whilst there is no right or wrong answer, but in general you want to start on the side where you have something to work with. Looking at the above question this is clearly the left side.
Even then the numerator is to the power 4, you can break this down into a difference of two squares which allows you to perform some further factorising.
\begin{aligned} & \equiv \frac{\cos ^4 \theta-\sin ^4 \theta}{\cos ^2 \theta} . \\ & \equiv \frac{\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\cos ^2 \theta} \\ & \equiv \frac{\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\cos ^2 \theta} \\ & \equiv \frac{\cos ^2 \theta}{\cos ^2 \theta}-\frac{\sin ^2 \theta}{\cos ^2 \theta} \\ & \equiv 1-\tan ^2 \theta=\text { RHS. } \end{aligned}Working with trig identities – example
Give p=3 \cos \theta \text {, and that } q=2 \sin \theta \text {, show that } 4 p^2+9 q^2=36 \text {. }
Here you are given two terms p and q and they are in terms of . The aim is to eliminate and to then use a trig identity to obtain the required expression as shown below:
p=3 \cos \theta \text {, and } q=2 \sin \theta \text {, }\cos \theta=\frac{p}{3} \text { and } \sin \theta=\frac{q}{2}Using \sin ^2 \theta+\cos ^2 \theta \equiv 1
\begin{aligned} \left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^2 & =1 \\ \frac{q^2}{4}+\frac{p^2}{9} & =1 \\ 4 p^2+9 q^2 & =36 \end{aligned}Proving trig identities – further examples
The following questions require the use of an appropriate trig identity in order to prove them. Try the questions on your own before you look at the solutions.
Part a)
(\sin \theta+\cos \theta)^2 \equiv 1+2 \sin \theta \cos \thetaPart a) solution:
Starting with the left side we have:
\begin{aligned} & =\sin ^2 \theta+2 \sin \theta \cos \theta+\cos ^2 \theta \\ & =\left(\sin ^2 \theta+\cos ^2 \theta\right)+2 \sin \theta \cos \theta \\ & =1+2 \sin \theta \cos \theta \end{aligned}Part b)
\frac{1}{\cos \theta}-\cos \theta \equiv \sin \theta \tan \thetaPart b) solution
Starting with the left side we have:
\begin{aligned} & =\frac{1-\cos ^2 \theta}{\cos \theta} \\ & =\frac{\sin ^2 \theta}{\cos \theta} \\ & =\sin \theta \times \frac{\sin \theta}{\cos \theta} \\ & =\sin \theta \tan \theta \end{aligned}Part c)
\tan x+\frac{1}{\tan x} \equiv \frac{1}{\sin x \cos x}Starting with the left side we have:
\begin{aligned} & =\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \\ & =\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x} \\ & =\frac{1}{\sin x \cos x} \end{aligned}Part d)
\cos ^2 A-\sin ^2 A \equiv 2 \cos ^2 A-1=1-2 \sin ^2 APart d) solution
Starting with the left side we have:
\begin{aligned} & \equiv \cos ^2 A-\left(1-\cos ^2 A\right) \\ & \equiv \cos ^2 A-1+\cos ^2 A \\ & \equiv 2 \cos ^2 A-1 \\ & \equiv 2\left(1-\sin ^2 A\right)-1 \\ & \equiv 2-2 \sin ^2 A-1 \\ & \equiv 1-2 \sin ^2 A \end{aligned}Here the question is asking you to obtain two answers which require the use of a trig identity.
Part e)
(2 \sin \theta-\cos \theta)^2+(\sin \theta+2 \cos \theta)^2 \equiv 5Part e) solution
\begin{aligned} \equiv & 4 \sin ^2 \theta-4 \sin \theta \cos \theta+\cos ^2 \theta \\ & +\sin ^2 \theta+4 \sin \theta \cos \theta+4 \cos ^2 \theta \\ \equiv & 5 \sin ^2 \theta+5 \cos ^2 \theta \\ \equiv & 5\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ \equiv & 5 \end{aligned}Remember there is no right or wrong approach when proving trig identities. But you need to make sure that you have something to work with. If we had started with the right side with this last question it would have been very difficult to proceed as there is no further information.
Part f)
2-(\sin \theta-\cos \theta)^2 \equiv(\sin \theta+\cos \theta)^2Part f) solution
Starting with the left we have:
\begin{aligned} & =2-\left(\sin ^2 \theta-2 \sin \theta \cos \theta+\cos ^2 \theta\right) \\ & =2-(1-2 \sin \theta \cos \theta) \\ & =1+2 \sin \theta \cos \theta \\ & =\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta \\ & =(\sin \theta+\cos \theta)^2 \end{aligned}Part g)
\sin ^2 x \cos ^2 y-\cos ^2 x \sin ^2 y \equiv \sin ^2 x-\sin ^2 yPart g) solution
\begin{aligned} = & \sin ^2 x\left(1-\sin ^2 y\right) \\ & -\left(1-\sin ^2 x\right) \sin ^2 y \\ = & \sin ^2 x-\sin ^2 x \sin ^2 y \\ & -\sin ^2 y+\sin ^2 x \sin ^2 y \\ = & \sin ^2 x-\sin ^2 y \end{aligned}You will notice that with all questions, we have started with the left and shown that it is identical to what is on the right. This is because there is more to work with on the left, when it comes to these questions.
Remember, proving trig identities is not always easy and there is no set method of which you can follow. It is always a case of trying something and hoping that it will work.