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## A Maths Tuition: Trigonometry Formulas

Many students find Trigonometry in A Maths challenging.

This is a list of Trigonometry Formulas that I compiled for A Maths. Students in my A Maths tuition class will get a copy of this, neatly formatted into one A4 size page for easy viewing.

A Maths: Trigonometry Formulas

$\mathit{cosec}x=\frac{1}{\sin x}$$\mathit{sec}x=\frac{1}{\cos x}$

$\cot x=\frac{1}{\tan x}$$\tan x=\frac{\sin x}{\cos x}$

(All Science Teachers Crazy)

$y=\sin x$

$y=\cos x$

$y=\tan x$

$\frac{d}{\mathit{dx}}(\sin x)=\cos x$

$\frac{d}{\mathit{dx}}(\cos x)=-\sin x$

$\frac{d}{\mathit{dx}}(\tan x)=\mathit{sec}^{2}x$

$\int {\sin x\mathit{dx}}=-\cos x+c$

$\int \cos x\mathit{dx}=\sin x+c$

$\int \mathit{sec}^{2}x\mathit{dx}=\tan x+c$

Special Angles:

$\cos 45^\circ=\frac{1}{\sqrt{2}}$

$\cos 60^\circ=\frac{1}{2}$

$\cos 30^\circ=\frac{\sqrt{3}}{2}$

$\sin 45^\circ=\frac{1}{\sqrt{2}}$

$\sin 60^\circ=\frac{\sqrt{3}}{2}$

$\sin 30^\circ=\frac{1}{2}$

$\tan 45^\circ=1$

$\tan 60^\circ=\sqrt{3}$

$\tan 30^\circ=\frac{1}{\sqrt{3}}$

$y=a\sin (\mathit{bx})+c$ Amplitude: $a$; Period: $\frac{2\pi }{b}$

$y=a\cos (\mathit{bx})+c$ Amplitude: $a$; Period: $\frac{2\pi }{b}$

$y=a\tan (\mathit{bx})+c$ Period: $\frac{\pi }{b}$

$\pi \mathit{rad}=180^\circ$

Area of  $\triangle \mathit{ABC}=\frac{1}{2}\mathit{ab}\sin C$

Sine Rule:  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Cosine Rule:  $c^{2}=a^{2}+b^{2}-2\mathit{ab}\cos C$

## Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$: area of $\triangle MXB$=9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$. (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means  that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.

Hence,

$\begin{array}{rcl} \triangle BDC &=& \frac{4}{3} \triangle BCN\\ &=& \frac{4}{3} (9u+6u)\\ &=& 20u \end{array}$

Thus, area of $ABCD=2 \triangle BDC=40u$

area of rectangle ABCD: area of $\triangle XBC$=40:6=20:3