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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}

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\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}}

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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.

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(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