Trigonometry Formula |
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sin^{2} A + cos^{2} A = 1 |

sec^{2}A + 1 = tan^{2} A |

cosec^{2}A = 1 + cot^{2}A |

sin(A±B) = sin A cos B ± cos A sin B |

cos(A±B) = cos A cos B ∓ sin A sin B |

tan(A±B) = ^{tan A ± tan B }⁄_{1∓tan A tan B} |

sin2A = 2 sinA cosA |

cos2A = cos^{2}A - sin^{2}A = 2cos^{2}A - 1 = 1 -2sin^{2}A |

tan2A = ^{2tan A }⁄_{1-tan2A} |

Proving trigonometries identities is a type of question that is struggled by many students. The common errors can stem from having serious algebraic misconceptions that they carry forward from lower secondary or they simply do not know the **proper techniques** to prove. More often than not, students will basically apply on whatever formulas they can see before their eyes, taking a big detour to get to the answer (if they manage to get to their destination). We should prescribe the right medicine for a particular illness and **cherry pick the formulas** by observation. All these will come with **consistent practices** and **good observation skills**. Students who are particularly good at proving may not have high intelligence, one thing we can be sure is that they have their definite share of hard work put in and had reaped their rewards. Another point that I would like to point out is to **MEMORISE ALL trigonometry formulas** and know them at the back of their hands for efficiency of the chapters. We do not have the luxury of staring at the formula list in examinations and take our own sweet time. Let us move on to these 2 questions from Chung Cheng High and Nanyang Girls High.

Rules:

- Start from the side that contains \(A \pm B\) (instead of \(AB\) or \(\frac{A}{B}\) ) or the side which looks more complicated.
- Change all trigonometry to SINE or COSINE (these 2 are the basic trigo)
- Factorise and/or Make into same denominator where necessary.
- Application of trigonometry formulas

Note that Rules 2, 3 and 4 are not in running orders and are still dependent on observation skills of the trigonometry present on both sides. However, these are tried and tested rules that work well most of the times based on years of experience. The exceptions I wont cover here as these are special cases. All my students can work pretty fast using the rules.

**Chung Cheng High (Yishun) Sec 4 Prelim 2017 P2/Q12**

**Example 1: **Prove the identity \(\left(\frac{1}{\sin \theta} – \frac{1}{\tan \theta} \right)^2 = \frac{1-\cos\theta}{1+\cos\theta}\).

LHS

\(=\left(\frac{1}{\sin \theta} – \frac{1}{\tan \theta} \right)^2\)

\(=\left(\frac{1}{\sin \theta} – \frac{1}{\left(\frac{\sin \theta}{\cos \theta} \right)} \right)^2 \)

\(=\left(\frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \right)^2\)

\(=\left(\frac{1-\cos\theta}{\sin \theta} \right)^2\)

\(=\frac{(1-\cos\theta)^2}{\sin^2\theta}\)

\(=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\)

\(=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\)

\(=\frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}\)

\(=\frac{1-\cos\theta}{1+\cos\theta}\)

=RHS

Voilà! We achieved the proof by making use of the standard rules.

**Nanyang Girls High 2017 Sec 4 MYE P2/Q10**

**Example 2**: Show that \(\frac{2\cos^2y+\cot y}{1+\text{cosec }2y}=2\cos^2y\).

LHS

\(=\frac{2\cos^2y+\cot y}{1+\text{cosec }2y}\)

\(=\frac{\left(2\cos^2y + \frac{\cos y}{\sin y}\right)}{1 + \frac{1}{\sin 2y}}\)

\(=\frac{\left(\frac{2\cos^2y\sin y+\cos y}{\sin y}\right)}{\left(\frac{\sin 2y+1}{\sin 2y}\right)}\)

\(=\frac{\cos y(2\sin y \cos y + 1)}{\sin y} \times \frac{\sin 2y}{\sin 2y+1}\)

\(=\frac{\cos y(\sin 2y + 1)}{\sin y} \times \frac{2\sin y \cos y}{\sin 2y + 1}\)

\(=2\cos^2y\)

=RHS

Remember the use of algebraic expansion in sec 2 whereby \((a+b)^2\) where most students will just expand it as \((a+b)(a+b)\). The truth is that students are expected to memorise \((a\pm b)^2 = a^2 \pm 2ab + b^2\) to have an edge over others as well as to shorten the presentation in examinations.

This is the bare minimum that students have to know before they move on to “Binomial Expansions” in sec 3 Additiional Mathematics.

In Binomial expansions, when the powers of binomials get unrealistically big, it is simply cumbersome to work out the slow expansions. Binomial Theorem then sets in where students can observe a guided pattern upon expansion.

To do well in this chapter, besides the Sec 2 foundation in ALGEBRA and Sec 3 INDICES, students need to be well versed in the observation and the techniques that can be applied to the different types of questions in Binomial Expansions.

For students who first join Math Academy and requested us to check their Amath script, it is common to see a handful of students who will just expand slowly in a desperate bid just to score that few marks in examinations. Hmm, did they know that they are eating into the precious time of other questions? Some will eventually find that needle in the haystack, but some will give up after a few attempts of algebraic errors. Well, there are always more efficient ways of getting things done.

Let me highlight using the example extracted from 2018 Nan Hua High Paper.

Q1. (i) By considering the **general term** in the binomial expansion of \(\left(x^2 + \frac{2}{x} \right)^{15}\), explain why all of the **powers of ** \(x\) in this expansion are **multiples of 3**. [3]

(ii) Find the coefficient of \(x^{22}\) in the expansion of \(\left(3x + \frac{1}{x^2} \right)\left(x^2 + \frac{2}{x} \right)^{15}\). [4]

(i)

Let us apply the general term formula: \(T_{r+1} = ^{15}C_r(x^2)^{15-r}\left(\frac{2}{x} \right)^r\)

We are only keen to know the powers of \(x\), hence we will pull out the terms that contain \(x\) as the base.

\((x^{30-2r})\left(\frac{1}{x} \right)^r\)

\(= \frac{x^{30-2r}}{x^r}\)

\(=x^{30-3r}\)

\(=x^{3(10-r)}\)

Since the power of \(x\) is \(3(10-r)\), which is a multiple of 3, therefore all of the **powers of ** \(x\) in this expansion are **multiples of 3.**

(ii)

For expansions of 2 or more brackets, most students will just expand blindly without further thought. That is a waste of precious time in examination. This is a 4 marks question and we have to complete this within a 6 minutes timeframe.

We have to **first observe the combination** that will pair up to get \(x^{22}\).

\(3x\) in the first bracket needs to pair with \(x^{21}\) from the second bracket and \(\frac{1}{x^2}\) in the first bracket needs to pair with \(x^{24}\) from the second bracket.

That is, \(\left(3x + \frac{1}{x^2} \right)\left( x^2 + \frac{2}{x}\right)^{15} =\left(3x + \frac{1}{x^2} \right)\left(Ax^{21} + Bx^{24} + \ldots \right)\)

We attempt to retrieve only \(x^{21}\) and \(x^{24}\).

From the general term in part (i),

\(x^{30-3x} = x^{21}\)

\(r=3\)

Substitute \(r=3\) into general term,

\(T_{3+1} = T_4\)

\(=^{15}C_3(2)^3x^{21}\)

\(=3640x^{21}\)

and

\(x^{30-3x} = x^{24}\)

\(r=2\)

Substitute \(r=2\) into general term,

\(T_{2+1} = T_3\)

\(=^{15}C_2(2)^2x^{24}\)

\(=420x^{24}\)

The expansion then becomes \(\left(3x + \frac{1}{x^2} \right)\left(3640x^{21} + 420x^{24} + \ldots \right)\)

Coefficient of \(x^{22} = (3 \times 3640) + (1 \times 420) = 11340\)

(MGS Prelim 2016 AMath Paper 2 Q 1b)

This question is in line with the 4047 Syllabus where students have to know

“Conditions for \(ax^2+bx+c\) to be always positive (or always negative)”

*Above extracted from SEAB.*

**Question**: Determine the conditions for \(p\) and \(q\) such that the curve \(y=px^2-2x+3q\) lies **entirely above** the\(x\)axis, where \(p\) and \(q\) are constants. [3 marks]

**Solution**: The keyword is “entirely above”, implying that the graph is above the axis. We expect a Happy Face that is not intersecting the x-axis.

\(\Rightarrow\) Happy Face is ** ABOVE** axis

\(\Rightarrow\) ** No** Real Solutions

\(\Rightarrow\) \(b^2-4ac < 0\)

Observe that \(a=p\), \(b=-2\) and \(c=3q\),

\(b^2-4ac < 0 \)

\((-2)^2 – 4(p)(3q) < 0 \)

\(4-12pq<0\)

\(1-3pq<0\)

\(\frac{1}{3} < pq \)

Since the graph is a happy face, we know that the coefficient of \(x^2\) is positive and \(p>0\).

Therefore, \( p>0\) and \(pq > \frac{1}{3} \)

]]>Let’s take a look at this A maths question on **Equations and Inequalities.** A handful of students will be confused with the keywords or how to effectively determine the correct sign of the discriminant.

The example below is very similar to **St Margaret 2016** A Math Prelim Paper 1 Q2.

Original Question: Find the range of values of \(p\) for which \((p+2)x^2 – 12x + 2(p-1)\) is always negative.

__Modified Example__: Find the range of values of \(p\) for which \((p+2)x^2 – 12x > -2(p-1)\) for all real values of \(x\).

**Misconception**:

Students will mistaken REAL as the keyword and likely think that \(b^2-4ac \geq 0\).

The emphasis is that the word REAL is crucial ONLY if coupled with a friend called ROOTS. We need to see the Keyword – **REAL ROOTS** for \(b^2-4ac \geq 0\) to hold true.

**Step 1:**

Ensure that the right hand side of the inequality is** zero**.

**Step 2:**

Having done so, recognise the keyword as “\(>0\)“. This implies that the graph will be above the axis and the shape of this quadratic expression will be a **Happy Face**. Since the U-shape is **not intersecting** with the axis, the solutions will be **imaginary** and \(b^2-4ac < 0\).

**Keyword **\(>0\) implies

\(\Rightarrow\) Happy Face is ** ABOVE** axis

\(\Rightarrow\) ** No** Real Solutions

\(\Rightarrow\) \(b^2-4ac < 0\)

**Step 3:**

Observe \(a=p+2, b=-12\) and \(c=2p-2\)

For imaginary solutions, we have :

\(b^2-4ac<0\)

\((-12)^2-4(p+2)(2p-2)<0\)

\(144-4(2p^2-2p+4p-4)<0\)

\(-8p^2-8p+160<0\)

\(-p^2-p+20<0\)

\((-p+4)(p+5)<0\)

\(p<-5\) or \( p>4\)…..(1)

At this stage, most students are contented with getting the solutions and will think that the question is completed.

Well, we are almost there. This is a **typical example** of having a **variable coefficient** (\(p+2\)) in front of \(x^2\).

__Extra Step is mandatory__

Remember we mentioned earlier that the graph is a Happy Face, thus we expect the coefficient of \(x^2\) to be positive and we have

\(p+2>0\)

\(p>-2\)….. (2)

Visualise the overlapping region of answers on a **number line **from (1) and (2),

Therefore \(p>4\) (final solution)

]]>

Increasingly asked in many different prelim questions, this type of question, which usually contains the keywords

The technique to solve this type of question is generic, as we will discuss in detail below.

Q4 (ii). The complex number \(z\) is given by $$z=-1+\text{i}c$$, where \(c\) is a non-zero real number.

Given that \(\frac{z^n}{z^*}\) is purely real, find the three smallest positive integer values of \(n\) when \(c=\sqrt{3}\).

**Solution with explanation:**

* Step 1: *To solve this type of question, we first find the

\(|z|=2\), \(\arg(z)=\frac{2\pi}{3}\)

* Step 2: *Convert to exponential form, then

Notice that by doing this, the power \(n\) is now “stuck” inside the trigo function, instead of being in the power. This is exactly what we want.

** Step 3: **For purely real, we set

Since\(\frac{z^n}{z^*}\) is purely real, Im(\(\frac{z^n}{z^*}\))\(=0\),

\(\sin\left[\left(n+1\right)\frac{2\pi}{3}\right]=0\)

**\((n+1)\frac{2\pi}{3}=k\pi, k \in \mathbb{Z}\)

\(n+1 = \frac{3k}{2}\)

We want the smallest 3 positive integers of \(n\), which corresponds to \(k=2,4,6\)

\(n+1=3, 6, 9\)

\(n = 2, 5, 8\)

**Note: We need to learn the generic solution for \(\sin \theta = 0\) and \(\cos \theta = 0\). To remember the following results, you just have to recall the graph of sine or cosine, and note down the positions of the zeros!

\(\sin \theta = 0\)

\(\theta =k\pi, k \in \mathbb{Z}\)

\(\cos \theta = 0\)

\(\theta =\frac{\pi}{2} + k\pi, k \in \mathbb{Z}\)