Part 2: Extension 1 Trigonometric Functions | Beginner’s Guide to Y11 Extension 1 Maths

Are you having a hard time with Extension 1 Trigonometric Functions? Well you’re in the right place. In this article, we’re going to explain how to satisfy all of the NESA syllabus outcomes for this topic!

In this article, we’ll discuss:

• The NESA Syllabus Outcomes
• The Compound Angle Formulae
• Sum-to-Product Formulae
• t-formulae
• Inverse Trigonometry
• Graphing Inverse Trigonometry
• Concept Check Questions

Year 11 Extension Trigonometric Functions

Mastering Trigonometric Functions is a very important skill in high school mathematics. By being able to understand and efficiently use harder formulae, students have an arsenal of methods to use when solving more complex questions. Further, grasping inverse functions provides a deeper insight into the nature of periodic functions, an essential tool in many modelling programs, etc.

NESA Syllabus Outcomes

NESA requires students to be proficient in the following syllabus outcomes:

• Derive and use the sum and difference expansions for the trigonometric functions \(\sin (? ± ?), \ \cos (? ± ?)\) and \(tan (? ± ?)\)
• Derive and use the double angle formulae for \(\sin 2?, \ \cos 2?\) and \(\tan 2?\) (ACMSM044)
• Derive and use expressions for \(\sin ?, \ \cos ?\) and \(\tan ?\) in terms of \(?\) where \(t = tan \frac{A}{2}\) (the ?-formulae)
• Derive and use the formulae for trigonometric products as sums and differences for \(\cos \ ? \ \cos \ ?, \ \sin?\sin?, \ \sin?\cos?\) and \(\cos?\sin?\) (ACMSM047)
• Define and use the inverse trigonometric functions (ACMSM119)
• Sketch graphs of the inverse trigonometric functions
• Use the relationships \(\sin(\sin^{-1} x=x\) and \(\sin^{-1} (\sin x)=x \ , \cos(\cos^{-1} x)=x\) and \(\cos^{-1} \cos x=x\), and \(\tan(\tan^{-1} x=x\) and \(\tan^{-1} (\tan x)=x\) where appropriate, and state the values of \(?\) for which these relationships are valid
• Prove and use the properties: \(sin^{-1} (-x)=-\sin^{-1} (x) \ , \cos^(-1) (-x)=π-\cos^{-1} (x) \ , \tan^{-1} (-x)=-\tan^(-1) (x)\) and \(\sin^{-1} (x)  +\cos^{-1} (x)=\frac{π}{2}\)
• Solve problems involving inverse trigonometric functions in a variety of abstract and practical situations AAM

Assumed Knowledge

Students should already by familiar with basic trigonometry concepts. This includes being proficient at radian measure and basic trigonometric function. Students can refresh their knowledge on basic trigonometric functions in the following Year 11 Subject Guide:

• Trigonometric Functions:

The Compound Angle Formulae

Before going through what the compound angle formulae are, we will first see what the compound formulae are not. This is one of the most common mistakes students make.

Common Mistakes

Note carefully:

\begin{align*}
sin (α+β)&≠sin⁡(α)+sin⁡(β)\\
cos⁡(α+β)&≠cos⁡(α)+cos⁡(β)\\
tan⁡(α+β)&≠tan⁡(α)+tan⁡(β)\\
\end{align*}

In fact, \(f(x+y)≠f(x)+f(y)\) for most functions \(f\). The probable (and understandable!) reason why students get this mixed up is because they are expanding as if it was algebra: \(a(b+c)=ab+ac\).

However, rules for functions are different to algebra. In fact, here are the actual Compound Angle Formulae.

The Compound Angle Formulae (For Sine and Cosine)

The following formulae are true for any real \(α\) and \(β\)

\begin{align*}
sin⁡(α+β)&=sin⁡(α)  cos⁡(β)+cos⁡(β)  sin⁡(α)\\
cos⁡(α+β)&=cos⁡(α)  cos⁡(β)-sin⁡(α)  sin⁡(β)\\
sin⁡(α-β)&=sin⁡(α)  cos⁡(β)-cos⁡(β)  sin⁡(α)\\
cos⁡(α-β)&=cos⁡(α)  cos⁡(β)+sin⁡(α)  sin⁡(β)\\
\end{align*}

Naturally, these formulae look very complex and possibly overwhelming. A good way to remember it is that:

\begin{align*}
\begin{cases}
&\sin \ ⟹\ \text{Same Sign, Different Function} \\
&\cos \ ⟹ \ \text{Different Sign, Same Function} \\
\end{cases}
\end{align*}

To explain, notice for the first compound angle formula, when you take sine of \(α\) plus \(β\), the formula has a plus in it (Same Sign), however it goes \(\sin \ \cos + \cos \ \sin\) (Different Function).

Try applying this logic to the remaining three formulas to notice the pattern!

The Compound Angle Formula (For Tangent)

The following formula is true for all (defined) real values of \(α, \ β\):

(where \(\cos⁡(α+β)≠0\))

Similarly,

Try to derive this from the compound angle formula for sin and cos!

Example 1.

Without using a calculator, find the exact value of \(\tan⁡(75°)\).

Solution 1:

Note that \(75°=30°+45°\), and both \(30°\) and \(45°\) have exact values when evaluated individually.

\begin{align*}
\tan⁡(75°)&=\tan⁡(30°+45° )\\
&=\frac{\tan⁡(30° )+\tan⁡(45° )}{1-\tan⁡(30° ) \tan⁡(45° ) }\\
&=\frac{(\frac{1}{\sqrt{3}})+(1)}{1-(\frac{1}{\sqrt{3}(1)}})\\
&=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
\end{align*}

where we multiplied the numerator and denominator by \(\sqrt{3}\) to get to the last step.

Note: an important special case of the Compound Angle Formulae are the Double Angle formulae.

The Double Angle Formulae

If you let \(α=β\) for \(\sin⁡(α+β)\) or \(\cos⁡(α+β)\) or \(tan⁡(α+β)\), you will get the double angle formulae.

\begin{align*}
\sin⁡(2α)&=2 \sin⁡(α)\cos⁡(α)\\
\tan⁡(2α)&=\frac{2 \tan⁡(α)}{1-\tan^2⁡(α)}\\
\end{align*}

There are three variations for the double angle formula for cosine (the second and third are derived from the first using \(\sin^2⁡(α)+\cos^2⁡(α)=1\))

\begin{align*}
\cos⁡(2α)&=\begin{cases}
&\cos^2⁡(α)-\sin^2⁡(α)\\
&2 \cos^2⁡(α)-1\\
&1-2 sin^2⁡(α)\\
\end{cases}
\end{align*}

Example 2.

Find the exact value of \(\sin⁡(120°)\) using the double angle formula.

Solution 2:

Note that \(\sin⁡(120° )=\sin⁡(2×60° )=2 \sin⁡(60°) \ cos⁡(60° )=2(\frac{\sqrt{3}}{2})(\frac{1}{2})=\frac{\sqrt{3}}{2}\).

We could have done this use the ASTC diagram as well, this example is just for illustrative purposes. See the Matrix Workbook or the Worksheet for further use of these formulae.

Sums and Products Formulae (For Sine and Cosine)

The sums and products formulae are usually seen in two forms:

Sums to products:

These formulae are easier to use if you have the sum of two trigonometric functions and you want to express them as a product:

\begin{align*}
\sin⁡(A)+\sin⁡(B)&=2 \sin⁡(\frac{A+B}{2}) \cos⁡(\frac{A-B}{2})\\
\sin⁡(A)-\sin⁡(B)&=2 \sin⁡(\frac{A-B}{2}) \cos⁡(\frac{A+B}{2})\\
\end{align*}

(The second formula comes directly from the first, since \(-\sin⁡(B)=\sin⁡(-B)\)).

Similarly

\begin{align*}
\cos⁡(A)+\cos⁡(B)&=2 \cos⁡({A+B}{2})  \cos⁡(\frac{A-B}{2})\\
\cos⁡(A)-\cos⁡(B)&=-2 \sin⁡({A+B}{2})  \sin⁡(\frac{A-B}{2})\\
\end{align*}

Products to Sums:

These formulae are easier to use when you have the product of two trigonometric functions and you want to express them as a sum.

\begin{align*}
\sin⁡(A) \sin⁡(B)&=\frac{1}{2} (\cos⁡(A-B)-\cos⁡(A+B) )\\
\cos⁡(A)  \cos⁡(B)&=\frac{1}{2} (\cos⁡(A-B)+\cos⁡(A+B) )\\
\sin⁡(A)  \cos⁡(B)&=\frac{1}{2} (\sin⁡(A+B)+\sin⁡(A-B) )\\
\end{align*}

These formulae may look daunting, but actually, you already know them! They are derived from the compound angle formula. Try to prove them yourself!

Example 1:

Express \(\sin⁡(\frac{π}{3})+\sin⁡(\frac{π}{6})\) as the product of trigonometric functions.

Solution 1:

Using the sums-to-product formula, we let \(A=\frac{π}{3}\) and \(B=\frac{π}{6}\), and we get

\begin{align*}
\sin⁡(\frac{π}{3})+\sin⁡(\frac{π}{6})  &=2 \sin⁡(\frac{\frac{π}{3}+\frac{π}{6}}{2}) \ cos⁡(\frac{\frac{π}{3}-\frac{π}{6}}{2})\\
&=2 \sin⁡(\frac{π}{4})  \sin⁡(\frac{π}{12})\\
\end{align*}

t – Formulae

The t- Formulae arise directly from the compound angle formulae, with a little more insight. They help with converting trigonometric functions into algebraic functions (of \(t\))– which are often far easier to work with!Let  \(t=tan⁡(\frac{x}{2})\). The t-formulae as follows:

\begin{align*}
\sin⁡(x)&=\frac{2t}{1+t^2} \\
\cos⁡(x)&=\frac{1-t^2}{1+t^2}\\
\tan⁡(x)&=\frac{2t}{1-t^2}\\
\end{align*}

Note that we can visualise this in a triangle:

[image 1]

Example 1.

Show that

using \(t\) formulae.

Solution 1:

Let \(t=tan⁡(\frac{x}{2})\). Then, we have expressions for \(\tan⁡(x)\) and \(\cos⁡(x)\) (and hence \(\sec⁡(x)\))

\begin{align*}
\tan⁡(x)&=\frac{2t}{1-t^2}\\
\cos⁡(x)&=\frac{1-t^2}{1+t^2}\\
\end{align*}

Hence,

\begin{align*}
“LHS”&=1+(\frac{2t}{1-t^2})^2\\
&=\frac{(1-t^2 )^2+4t^2}{(1-t^2 )^2}\\
&=\frac{t^4+2t^2+1}{(1-t^2 )^2}\\
&=\frac{(1+t^2 )^2}{(1-t^2 )^2}\\
&=(\frac{1+t^2}{1-t^2 })^2\\
&=\sec^2⁡(x)=”RHS”\\
\end{align*}

Inverse Trigonometry

Recall from inverse functions topic, when functions do not have one-to-one correspondence, we can restrict the domain of the function and define the inverse function over that smaller domain.

We do this for all three trigonometric functions – \(\sin⁡(x)\), \(\cos⁡(x)\) and \(\tan⁡(x)\).

• We restrict \(\sin⁡(x)\) to the domain \(x∈[-\frac{π}{2},\frac{π}{2}]\)
• We restrict \(\cos⁡(x)\) to the domain \(x∈[0,π]\)
• We restrict \(tan⁡(x)\) to the domain \(x∈(-\frac{π}{2},\frac{π}{2})\), not including the endpoints since tan is undefined at \(-\frac{π}{2}\) and \(\frac{π}{2}\).

The following table summarises the inverse functions, along with their domain and range:

Note: Ensure you always use radians!

Relationship between inverse sine and inverse cosine:

A very important result to know is that for \(x∈[0,\frac{π}{2}],\sin^{-1}⁡(x)+\cos^{-1}⁡(x)=\frac{π}{2}\)

Example 1.

Without using a calculator, find the exact value of \(\sin^{-1}⁡(\sin(-\frac{π}{6} ))\).

Solution 1:

Note: \(f^{-1} (f(x))=x\) for all \(x∈ \)(“restricted” )  “domain of” \(f\) is a key property to remember. In this case where \(f(x)=sin⁡(x)\) we have that \(-\frac{π}{6}∈[-\frac{π}{2},\frac{π}{2}\) and hence applying the result we get

\(\sin^{-1}⁡(\sin⁡(\frac{-π}{6}) )=-\frac{π}{6}\).

Sketching Inverse Functions

You should also be able to graph the inverse trigonometric functions. If you ever forget, you can draw the standard trigonometric functions for the restricted domain, and then reflect them about the line \(y=x\).

Inverse Function	Graph
\(sin^{-1}⁡(x)\)
\(cos^{-1}⁡(x)\)
\(tan^{-1}⁡(x)\)

As with other functions, we can sketch variations of these graphs by scaling and translation as covered in the Graphical Transformations topic.

Example 1.

Sketch \(y=\sin^{-1}⁡(2x)+3\).

Solution 1:

Note, the first thing to do is to determine the new domain after squashing the graph horizontally.

\begin{align*}
-\frac{π}{2}&≤2x≤\frac{π}{2}\\
-\frac{π}{4}&≤x≤\frac{π}{4}\\
\end{align*}

The \(+3\) as the effect of shifting the graph up by \(3\) units. Hence, the overall graph will look like this:

Concept Check Questions:

1. Find the exact value of \(\sin⁡(105°)\)
2. Show that \(\sin⁡(x+\frac{π}{4})=\frac{1}{\sqrt{2}} (sin⁡(x)+cos⁡(x) )\) and hence solve the equation
   \(\sin⁡(x)+\cos⁡(x)=1\)
   for \(x∈[0,2π]\).
3. Show that
   \(\frac{\sin^2⁡(x)}{1+cos⁡(x)}=1-cos⁡(x)\)
   using t formulae.
4. Without using a calculator, find the value of \(\sin^{-1}⁡(\sin⁡(\frac{4π}{3} ))\).
5. Sketch the graph of \(y=1-2 tan^{-1}⁡(3x-1)\)

Concept Check Solutions:

1. \(sin⁡(105°)=\frac{1+\sqrt{3}}{\frac{2}{\sqrt{2}}}\).
2. \(x=0, \ \frac{π}{2}\) or \(2π\).
3. NA
4. \(-\frac{π}{3}\)
5. 

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