Revision: trigonometry, exponentials, logarithms and sequences

These grade 11 topics rarely form one long question, but they are scattered all over the paper. Part I usually has two to four short questions: a trigonometric identity, a logarithmic equation, a term of a sequence, a limit. Part II or Part III often has a real-life model using an exponential or trigonometric function. Altogether these topics are usually worth about $1$ to $1.5$ marks.

This revision lesson recalls the formulas of each area briefly, then works examples exactly like exam questions.

Trigonometric functions and equations

Theorem Identities and basic equations

$\sin^2x+\cos^2x=1$ ; $\cos2x=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1$ ; $\sin2x=2\sin x\cos x$ .
$\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b$ ; $\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b$ .
$\sin x=\sin\alpha\iff x=\alpha+k2\pi$ or $x=\pi-\alpha+k2\pi$ ; $\cos x=\cos\alpha\iff x=\pm\alpha+k2\pi$ ; $\tan x=\tan\alpha\iff x=\alpha+k\pi$ ( $k\in\mathbb{Z}$ ).
The functions $y=A\sin(\omega x+\varphi)$ and $y=A\cos(\omega x+\varphi)$ have period $\dfrac{2\pi}{|\omega|}$ and take values in $[-|A|, |A|]$ .

When a question asks for the number of solutions in an interval, write the general solution and try each integer $k$ , or look at the unit circle.

Example Multiple choice: a basic cosine equation

The solutions of the equation $2\cos x+1=0$ are

A. $x=\pm\dfrac{2\pi}{3}+k2\pi$ , $k\in\mathbb{Z}$ .

B. $x=\pm\dfrac{\pi}{3}+k2\pi$ , $k\in\mathbb{Z}$ .

C. $x=\dfrac{2\pi}{3}+k\pi$ , $k\in\mathbb{Z}$ .

D. $x=\pm\dfrac{2\pi}{3}+k\pi$ , $k\in\mathbb{Z}$ .

Solution. We write it in the form $\cos x=\cos\alpha$ , using $\cos\dfrac{2\pi}{3}=-\dfrac12$ :

$2\cos x+1=0\iff\cos x=-\frac12=\cos\frac{2\pi}{3}\iff x=\pm\frac{2\pi}{3}+k2\pi.$

The answer is A. Option B corresponds to $\cos x=\dfrac12$ (a lost minus sign); options C and D wrongly use the period $\pi$ of $\tan$ .

Trigonometric functions model repeated motion very well, for example a capsule on an observation wheel.

Height of a capsule

h(t)=22-20\cos\dfrac{\pi t}{10}

over one turn

Example True/false: an observation wheel

An observation wheel turns at a constant speed. The height (in metres) of a capsule above the ground after $t$ minutes is $h(t)=22-20\cos\dfrac{\pi t}{10}$ , $t\ge0$ .

a) The lowest height of the capsule is $0$ m.

b) One turn of the wheel takes $20$ minutes.

c) At time $t=10$ minutes the capsule is $42$ m high.

d) During one turn, the capsule spends $10$ minutes at a height of $32$ m or more.

Solution. We use $-1\le\cos\dfrac{\pi t}{10}\le1$ and the period formula.

a) False. When $\cos\dfrac{\pi t}{10}=1$ , the lowest height is $22-20=2$ m.

b) True. The period is $\dfrac{2\pi}{\pi/10}=20$ minutes.

c) True. $h(10)=22-20\cos\pi=22+20=42$ m.

d) False. For $t\in[0, 20]$ we solve the inequality:

$22-20\cos\frac{\pi t}{10}\ge32\iff\cos\frac{\pi t}{10}\le-\frac12\iff\frac{2\pi}{3}\le\frac{\pi t}{10}\le\frac{4\pi}{3}\iff\frac{20}{3}\le t\le\frac{40}{3}.$

That time is $\dfrac{40}{3}-\dfrac{20}{3}=\dfrac{20}{3}\approx6.67$ minutes, not $10$ minutes.

So the answers are a) False, b) True, c) True, d) False.

Exponential and logarithmic functions

Theorem Powers, logarithms and models

For $0<a\neq1$ and $b, c>0$ : $\log_a(bc)=\log_ab+\log_ac$ , $\log_a\dfrac bc=\log_ab-\log_ac$ , $\log_ab^\alpha=\alpha\log_ab$ , $\log_ab=\dfrac{\log_cb}{\log_ca}$ .
$a^x=b\iff x=\log_ab$ (needs $b>0$ ); $\log_ax=b\iff x=a^b$ (needs $x>0$ ).
In inequalities: with $a>1$ keep the direction; with $0<a<1$ reverse it.
Compound interest: $A=A_0(1+r)^n$ . Growth or decay: $N(t)=N_0e^{kt}$ or $N(t)=N_0\cdot q^t$ (growth when $q>1$ , decay when $0<q<1$ ).

With a logarithmic equation, the first job is always to state the conditions. Skipping this step is the most common way to lose marks.

Example Short answer: a drug in the blood

After an injection, the concentration of a drug in a patient's blood is $C(t)=40\cdot0.8^t$ (mg/L), where $t$ is the number of hours since the injection. After how many hours has the concentration fallen to $5$ mg/L (to 1 decimal place)?

Solution. We solve the exponential equation by taking logarithms:

$40\cdot0.8^t=5\iff0.8^t=\frac18\iff t=\log_{0.8}\frac18=\frac{\ln0.125}{\ln0.8}\approx9.32.$

So after about $9.3$ hours (write $9.3$ ).

Check. Each hour $80\%$ of the drug remains, so after $9$ hours there is $40\cdot0.8^9\approx5.4$ mg/L, slightly more than $5$ ; the answer $9.3$ is sensible.

Arithmetic and geometric sequences

Theorem The two kinds of sequence

	Arithmetic sequence (common difference $d$ )	Geometric sequence (common ratio $q$ )
General term	$u_n=u_1+(n-1)d$	$u_n=u_1\cdot q^{n-1}$
Sum of the first $n$ terms	$S_n=\dfrac{n(u_1+u_n)}{2}=\dfrac{n\big[2u_1+(n-1)d\big]}{2}$	$S_n=\dfrac{u_1(1-q^n)}{1-q}$ ( $q\neq1$ )
Three consecutive terms	$u_k=\dfrac{u_{k-1}+u_{k+1}}{2}$	$u_k^2=u_{k-1}\cdot u_{k+1}$

Many real-life problems are sequences "in disguise": seats per row increasing steadily form an arithmetic sequence; money growing by a percentage forms a geometric sequence.

Limits

Key terms

English	Tiếng Việt
basic trigonometric equation	phương trình lượng giác cơ bản
period	chu kì
natural logarithm	lôgarit tự nhiên
compound interest	lãi kép
exponential growth	tăng trưởng mũ
arithmetic sequence	cấp số cộng
geometric sequence	cấp số nhân
common difference	công sai
common ratio	công bội
convergent geometric series (sum to infinity)	cấp số nhân lùi vô hạn
limit	giới hạn

Exercises

The exercises go from easy to hard, and each one states its exam format.

Exercise 1 Multiple choice

Given $\sin x=\dfrac13$ , the value of $\cos2x$ is

A. $\dfrac79$ . B. $\dfrac13$ . C. $-\dfrac79$ . D. $\dfrac89$ .

What you need and how to start

What you need: $\cos2x=1-2\sin^2x$ .

Steps:

Find $\sin^2x$ .
Substitute into the identity.

Show solution

$\cos2x=1-2\cdot\dfrac19=\dfrac79$ . Option B forgets to square, option C uses $2\sin^2x-1$ by mistake, and option D is $\cos^2x$ .

Answer. A.

Exercise 2 Multiple choice

The solutions of the equation $\tan x=\sqrt3$ are

A. $x=\dfrac{\pi}{3}+k\pi$ . B. $x=\dfrac{\pi}{3}+k2\pi$ . C. $x=\dfrac{\pi}{6}+k\pi$ . D. $x=\pm\dfrac{\pi}{3}+k\pi$ ( $k\in\mathbb{Z}$ ).

What you need and how to start

What you need: $\tan x=\tan\alpha\iff x=\alpha+k\pi$ ; $\tan\dfrac{\pi}{3}=\sqrt3$ .

Steps:

Find $\alpha$ with $\tan\alpha=\sqrt3$ .
Write the general solution with period $\pi$ .

Show solution

$\tan x=\tan\dfrac{\pi}{3}\iff x=\dfrac{\pi}{3}+k\pi$ . Option B misses half of the solutions; option C uses $\tan\dfrac{\pi}{6}=\dfrac{\sqrt3}{3}$ .

Answer. A.

Exercise 3 Multiple choice

Let $a=\log_23$ . Then $\log_224$ equals

A. $3a$ . B. $3+a$ . C. $8+a$ . D. $1+3a$ .

What you need and how to start

What you need: $\log_2(bc)=\log_2b+\log_2c$ .

Steps:

Write $24=8\cdot3$ .
Use $\log_28=3$ .

Show solution

$\log_224=\log_28+\log_23=3+a$ . Option A turns the addition into a multiplication; option C forgets to evaluate $\log_28$ .

Answer. B.

Exercise 4 True/false

The geometric sequence $(u_n)$ has $u_1=3$ and common ratio $q=2$ .

a) $u_5=48$ .

b) The sum of the first $5$ terms is $93$ .

c) $u_n=3\cdot2^n$ for every $n\ge1$ .

d) The number $384$ is the $7$ th term of the sequence.

What you need and how to start

What you need: $u_n=u_1q^{n-1}$ ; $S_n=\dfrac{u_1(1-q^n)}{1-q}$ .

Steps:

Compute $u_5$ and $S_5$ .
Solve $3\cdot2^{n-1}=384$ .

Show solution

a) True. $u_5=3\cdot2^4=48$ .

b) True. $S_5=\dfrac{3(1-2^5)}{1-2}=3\cdot31=93$ .

c) False. $u_n=3\cdot2^{n-1}$ ; the formula $3\cdot2^n$ gives $u_1=6$ .

d) False. $3\cdot2^{n-1}=384\iff2^{n-1}=128\iff n=8$ , so $384$ is the $8$ th term.

Answer. a) True; b) True; c) False; d) False.

Exercise 5 Short answer

Find the sum of the solutions of the equation $4^x-6\cdot2^x+8=0$ .

What you need and how to start

What you need: $4^x=(2^x)^2$ ; put $t=2^x>0$ .

Steps:

Solve the quadratic equation in $t$ .
Find $x$ from $2^x=t$ .

Show solution

$t^2-6t+8=0\iff t=2$ or $t=4$ (both positive). Then $x=1$ or $x=2$ , and the sum is $3$ .

Answer. $3$ .

Exercise 6 Short answer

How many integer solutions does the inequality $\log_{\frac12}(x-2)\ge-2$ have?

What you need and how to start

What you need: the base $\dfrac12<1$ , so the direction reverses; the condition is $x-2>0$ .

Steps:

Write $-2=\log_{\frac12}4$ .
Solve $0<x-2\le4$ .

Show solution

$\log_{\frac12}(x-2)\ge\log_{\frac12}4\iff0<x-2\le4\iff2<x\le6$ . The integers are $3$ , $4$ , $5$ , $6$ .

Answer. $4$ .

Exercise 7 Short answer

Your friend An deposits $100$ million đồng in a bank at $6\%$ interest per year, with the interest added to the balance every year. After at least how many years is the balance $150$ million đồng or more?

What you need and how to start

What you need: $A=A_0(1+r)^n$ ; $n$ is a positive integer.

Steps:

Solve $100\cdot1.06^n\ge150$ .
Take the least integer that works.

Show solution

$1.06^n\ge1.5\iff n\ge\log_{1.06}1.5\approx6.96$ . Check: $100\cdot1.06^6\approx141.9$ and $100\cdot1.06^7\approx150.4$ (million đồng).

Answer. $7$ .

Exercise 8 Short answer

A culture starts with $1000$ bacteria, and the number doubles every $3$ hours, so after $t$ hours there are $N(t)=1000\cdot2^{\frac t3}$ bacteria. After at least how many whole hours does the number exceed $20\,000$ ?

What you need and how to start

What you need: $2^{\frac t3}>20\iff\dfrac t3>\log_220$ .

Steps:

Solve the exponential inequality.
Take the least integer that works, then check.

Show solution

$t>3\log_220\approx12.97$ . Check: $N(12)=1000\cdot2^4=16\,000$ and $N(13)\approx20\,159>20\,000$ .

Answer. $13$ .

Exercise 9 Written answer

Solve the equation $\sin2x=\cos x$ on $[0, \pi]$ .

What you need and how to start

What you need: $\sin2x=2\sin x\cos x$ ; turn it into a product equal to zero.

Steps:

Write $\cos x(2\sin x-1)=0$ .
Solve each factor and keep the solutions in $[0, \pi]$ .

Show solution

$2\sin x\cos x-\cos x=0\iff\cos x(2\sin x-1)=0$ . On $[0, \pi]$ : $\cos x=0\iff x=\dfrac{\pi}{2}$ ; $\sin x=\dfrac12\iff x=\dfrac{\pi}{6}$ or $x=\dfrac{5\pi}{6}$ .

Answer. $x\in\left\{\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}\right\}$ .

Exercise 10 Short answer (challenge)

A ball is dropped from a height of $10$ m onto the floor. Each time it hits the floor, it bounces back up to $\dfrac34$ of its previous height. Assuming it bounces like this for ever, find the total distance travelled by the ball (in m).

What you need and how to start

What you need: the sum to infinity $S=\dfrac{u_1}{1-q}$ .

Steps:

The distance is the first drop of $10$ m, then for each bounce one trip up and an equal trip down.
The bounce heights $7.5$ , $7.5\cdot\dfrac34$ , … form a convergent geometric series.

Show solution

The bounce heights add up to $\dfrac{7.5}{1-\frac34}=30$ m. Each height is travelled twice, so the distance is $10+2\cdot30=70$ m.

Answer. $70$ .

Trigonometric functions and equations

Exponential and logarithmic functions

Arithmetic and geometric sequences

Limits

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The domain of the function is $(0, +\infty)$ .	True	False
b) The function is increasing on its domain.	True	False
c) The graph passes through the point $(0, 1)$ .	True	False
d) $\log_2x<3\iff x<8$ .	True	False