Graphs of sin x, cos x, tan x — key features, period, amplitude: Question bank

True or false — justify

Every statement below is either true or false. Decide, then give the reason.

sin x

can equal

1.2

for some large

x

False —

sin x

is a coordinate on a radius-

1

circle, so it can never leave

[- 1, 1]

no matter how large

x

grows; the wave just repeats, it never climbs higher.

cos x

and

sin x

are never equal for any

x

False — they are equal wherever the point sits on the diagonal

y = x

, e.g. at

x = \frac{π}{4}

where both equal

\frac{1}{2}

; see Exact Values of Trig Ratios.

The period of

tan x

is smaller than the period of

sin x

True —

tan

repeats every

π

while

sin

repeats every

2 π

, because the two sign-flips at

x + π

cancel in the ratio

sin / cos

tan x

has amplitude

1

because it is a trig function

False — near each asymptote

tan x

races off to

\pm \infty

, so it has no maximum or minimum; amplitude only exists for a bounded up-and-down swing.

cos x

is just

sin x

slid to the right by

\frac{π}{2}

False in direction — cosine leads sine (it peaks a quarter-turn earlier), so

cos x = sin (x + \frac{π}{2})

, a slide to the left.

sin (- x) = - sin x

but

cos (- x) = cos x

True — reflecting the angle across the horizontal axis flips the

y

-coordinate (sine, odd) but leaves the

x

-coordinate (cosine, even) untouched.

Every zero of

sin x

is also a place where

tan x = 0

True —

tan x = sin x / cos x

is zero exactly when its top

sin x

is zero and its bottom

cos x \neq = 0

; at

x = nπ

cosine is

\pm 1 \neq = 0

, so all these are genuine zeros.

Every zero of

cos x

is a zero of

tan x

False — where

cos x = 0

we are dividing by zero, so

tan x

has an asymptote there, the opposite of a zero.

The graph of

y = tan x

is continuous everywhere

False — it breaks at every

x = \frac{π}{2} + nπ

where cosine vanishes; between consecutive asymptotes it is one unbroken branch, but the asymptotes chop it into separate pieces.

Spot the error

Each line states a claim with faulty reasoning; the reveal names the flaw.

y = 3 sin (2 x)

has period

2 π \times 2 = 4 π

Error — a bigger

B

makes the wave faster, so you divide: period

= 2 π /2 = π

. Ask "how much must

x

grow so the inside grows by

2 π

y = tan (3 x)

has period

2 π /3

Error —

tan

's base period is

π

, not

2 π

, so the period is

π /3

; using

2 π

double-counts because the ratio already repeats after half a turn.

y = 5 cos x

has amplitude

- 5

when the cosine dips negative."

Error — amplitude is a size, always

∣ A ∣ = 5

; the

cos

going negative is already built into the

[- 5, 5]

range, not a change of amplitude.

sin x = 0

has no solutions because a wave never touches a line."

Error —

sin x

crosses the axis every half-period, at

x = 0, π, 2 π, \dots

; a smooth wave crosses zero, it does not avoid it.

"The midline of

y = 2 sin x + 3

y = 0

Error — the

+ 3

lifts the whole wave, so the midline is

y = 3

; the wave then swings between

1

and

5

. See Transformations of Graphs.

cos x

starts at

0

sin x

Error — at

x = 0

the point is at the right

(1, 0)

, so

cos 0 = 1

(a crest) while

sin 0 = 0

; cosine crests, sine starts at sea-level.

"To find the period of

y = sin (B x + C)

you must know

C

Error —

C

only slides the graph sideways (phase shift), it never changes how often it repeats; period depends on

B

alone,

2 π /∣ B ∣

tan x

can be made to have amplitude

4

by writing

y = 4 tan x

Error — multiplying by

4

stretches the curve vertically but it still shoots to

\pm \infty

; a function with no bound cannot acquire an amplitude.

Why questions

Answer with the mechanism, not just the fact.

Why is the range of

sin x

and

cos x

exactly

[- 1, 1]

Because both are coordinates of a point pinned to a circle of radius

1

; a coordinate can be at most the radius in size, so it lives between

- 1

and

1

Why does

tan x

have period

π

and not

2 π

Moving half a turn (

x \to x + π

) flips the point to the opposite side, negating both

sin

and

cos

; in the ratio

\frac{- sin x}{- cos x}

the two minus signs cancel, so it repeats after only

π

Why does

tan x

blow up to infinity at certain points?

x = \frac{π}{2} + nπ

the point is at the top or bottom of the circle, so its horizontal coordinate

cos x

0

; dividing

sin x

by that zero drives

tan x

off to

\pm \infty

Why does increasing

B

sin (B x)

shrink the period?

The inside

B x

reaches

2 π

sooner when

B

is bigger, so one full cycle is completed after a smaller change in

x

: period

= 2 π /∣ B ∣

Why is

cos x

even but

sin x

odd?

Negating the angle reflects the point across the horizontal axis; the horizontal position (

cos

) is unchanged (even) while the height (

sin

) flips sign (odd).

Why can we solve

tan x = 0

by solving

sin x = 0

A fraction is zero only when its numerator is zero, so

tan x = sin x / cos x = 0

needs

sin x = 0

— provided

cos x \neq = 0

so we are not at an asymptote. See Solving Trigonometric Equations.

Why do sine and cosine graphs describe a swinging pendulum or spring?

Their smooth, bounded, repeating up-and-down is exactly the motion of an object oscillating about a rest point, which is why

sin

and

cos

model Simple Harmonic Motion.

Why does the identity

tan x = sin x / cos x

warn you where

tan

misbehaves?

Reading it as a ratio makes the danger visible: wherever the denominator

cos x

hits zero, the whole expression is undefined; the identity is from Trigonometric Identities.

Edge cases

The scenarios that break naive rules.

What is

tan x

x = \frac{π}{2}

It is undefined —

cos (\frac{π}{2}) = 0

, so the ratio

sin / cos

divides by zero; the graph has an asymptote, not a value, there.

Does

tan x

take the value

0

and the value

\pm \infty

at the same places?

No — zeros happen where

sin x = 0

(at

x = nπ

) and asymptotes where

cos x = 0

(at

x = \frac{π}{2} + nπ

); these two sets never overlap, they alternate.

What is the amplitude of

y = 0 \cdot sin x

Zero — the wave is flattened onto the line

y = 0

; it is a degenerate constant function, technically periodic but with no swing at all.

What happens to the period of

y = sin (B x)

B \to 0

The period

2 π /∣ B ∣

grows without bound, so the wave stretches flatter and flatter; in the limit

B = 0

it becomes the constant

sin 0 = 0

with no repeating at all.

At exactly which points do

sin x

and

cos x

momentarily agree in value on

[0, 2 π]

x = \frac{π}{4}

(both

+ \frac{1}{2}

) and

x = \frac{5 π}{4}

(both

- \frac{1}{2}

), the two crossing points of their graphs.

Is the largest value of

tan x

on the open interval

(- \frac{π}{2}, \frac{π}{2})

ever reached?

No maximum exists — as

x \to \frac{π}{2}

from below,

tan x

increases without bound; it approaches

+ \infty

but never attains a top value.

Connections

Parent: the three graphs
Unit Circle and Radian Measure — the picture behind every trap here
Exact Values of Trig Ratios — supplies the crossing values like $\frac{1}{2}$
Trigonometric Identities — why $tan = sin / cos$ exposes the asymptotes
Transformations of Graphs — the $A, B, C, D$ traps
Solving Trigonometric Equations — zeros vs asymptotes in practice
Simple Harmonic Motion — where these waves come alive

#flashcards/maths