Intuition The big pictureEvery "wavy" sine graph you'll ever meet is just the basic sin x \sin x sin x stretched, squished, slid, and lifted . The four numbers A , B , C , D A, B, C, D A , B , C , D one of those actions. Learn what each does and why , and you can sketch any sinusoid in
seconds — no memorising ugly pictures.
WHY four numbers? A general wave needs to say: how tall (A), how fast it repeats (B),
where it starts (C), and what level it wobbles around (D). These are the four independent
freedoms of a wave, nothing more.
We start from y = sin x y=\sin x y = sin x 2 π 2\pi 2 π y = 0 y=0 y = 0
Replace y y y y / A y/A y / A y = A sin x y = A\sin x y = A sin x sin \sin sin [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] A A A [ − A , A ] [-A, A] [ − A , A ]
sin \sin sin input runs through 2 π 2\pi 2 π B x Bx B x B x increases by 2 π ⇒ Δ ( B x ) = 2 π ⇒ B Δ x = 2 π . Bx \text{ increases by } 2\pi \;\Rightarrow\; \Delta(Bx)=2\pi \;\Rightarrow\; B\,\Delta x = 2\pi. B x increases by 2 π ⇒ Δ ( B x ) = 2 π ⇒ B Δ x = 2 π . x x x
Factor the input: B x + C = B ( x + C B ) Bx + C = B\!\left(x + \tfrac{C}{B}\right) B x + C = B ( x + B C ) x x x x + C B x+\tfrac{C}{B} x + B C left by C B \tfrac{C}{B} B C + + + opposite way).
Adding D D D D D D midline y = D y=D y = D y = D − ∣ A ∣ y = D-|A| y = D − ∣ A ∣ y = D + ∣ A ∣ y = D+|A| y = D + ∣ A ∣
Intuition Do horizontals, then verticalsInside the sine (B B B C C C x x x Outside the sine (A A A D D D y y y = 2 π / ∣ B ∣ =2\pi/|B| = 2 π /∣ B ∣ = − C / B =-C/B = − C / B = D =D = D = ∣ A ∣ =|A| = ∣ A ∣
Worked example Example 1 —y = 3 sin ( 2 x − π 2 ) + 1 y = 3\sin\!\left(2x - \tfrac{\pi}{2}\right) + 1 y = 3 sin ( 2 x − 2 π ) + 1 Amplitude: ∣ A ∣ = 3 |A| = 3 ∣ A ∣ = 3 Why? A = 3 A=3 A = 3
Period: T = 2 π ∣ B ∣ = 2 π 2 = π T=\dfrac{2\pi}{|B|}=\dfrac{2\pi}{2}=\pi T = ∣ B ∣ 2 π = 2 2 π = π Why? Input 2 x 2x 2 x 2 π 2\pi 2 π x x x
Phase shift: factor 2 x − π 2 = 2 ( x − π 4 ) 2x-\tfrac{\pi}{2}=2\!\left(x-\tfrac{\pi}{4}\right) 2 x − 2 π = 2 ( x − 4 π ) + π 4 +\tfrac{\pi}{4} + 4 π Why? − C / B = − ( − π / 2 ) / 2 = π / 4 -C/B = -(-\pi/2)/2 = \pi/4 − C / B = − ( − π /2 ) /2 = π /4 ( x − π 4 ) (x-\tfrac\pi4) ( x − 4 π ) right slide.
Vertical: midline y = 1 y=1 y = 1 = 1 + 3 = 4 =1+3=4 = 1 + 3 = 4 = 1 − 3 = − 2 =1-3=-2 = 1 − 3 = − 2 Why? D = 1 D=1 D = 1 ∣ A ∣ |A| ∣ A ∣
Worked example Example 2 —y = − 2 cos ( π x ) + 5 y = -2\cos\!\left(\pi x\right) + 5 y = − 2 cos ( π x ) + 5 Amplitude: ∣ − 2 ∣ = 2 |-2| = 2 ∣ − 2∣ = 2 minimum , not maximum).
Why the flip? A < 0 A<0 A < 0
Period: T = 2 π ∣ π ∣ = 2 T = \dfrac{2\pi}{|\pi|} = 2 T = ∣ π ∣ 2 π = 2 Why? B = π B=\pi B = π x x x 2 2 2
Midline / max / min: midline y = 5 y=5 y = 5 = 7 =7 = 7 = 3 =3 = 3
At x = 0 x=0 x = 0 y = − 2 cos 0 + 5 = − 2 + 5 = 3 y=-2\cos 0+5 = -2+5 = 3 y = − 2 cos 0 + 5 = − 2 + 5 = 3
Worked example Example 3 — Reverse-engineer from a graphA wave has max = 8 =8 = 8 = 2 =2 = 2 4 4 4 x = 1 x=1 x = 1 A sin ( B x + C ) + D A\sin(Bx+C)+D A sin ( B x + C ) + D
D = max + min 2 = 8 + 2 2 = 5 D = \frac{\max+\min}{2} = \frac{8+2}{2}=5 D = 2 m a x + m i n = 2 8 + 2 = 5 Why? Midline is the average of top and bottom.A = max − min 2 = 8 − 2 2 = 3 A = \frac{\max-\min}{2} = \frac{8-2}{2}=3 A = 2 m a x − m i n = 2 8 − 2 = 3 Why? Half the total peak-to-trough height.B = 2 π T = 2 π 4 = π 2 B = \frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2} B = T 2 π = 4 2 π = 2 π Why? Invert the period formula.Starts up through midline at x = 1 x=1 x = 1 C = − B ⋅ ( shift ) = − π 2 C = -B\cdot(\text{shift}) = -\tfrac{\pi}{2} C = − B ⋅ ( shift ) = − 2 π Answer: y = 3 sin ( π 2 x − π 2 ) + 5. y = 3\sin\!\left(\tfrac{\pi}{2}x - \tfrac{\pi}{2}\right)+5. y = 3 sin ( 2 π x − 2 π ) + 5.
Recall Predict before you computeFor y = 4 sin ( 3 x + π ) − 1 y=4\sin(3x+\pi)-1 y = 4 sin ( 3 x + π ) − 1 forecast : amplitude? period? shift? midline?
Then verify below.
Worked example VerificationAmplitude = 4 =4 = 4 = 2 π / 3 =2\pi/3 = 2 π /3 = − π / 3 =-\pi/3 = − π /3 π / 3 \pi/3 π /3 3 x + π = 3 ( x + π / 3 ) 3x+\pi=3(x+\pi/3) 3 x + π = 3 ( x + π /3 ) y = − 1 y=-1 y = − 1 = 3 =3 = 3 = − 5 =-5 = − 5
Common mistake "Period is2 π B 2\pi B 2 π B Why it feels right: bigger B B B feels like a bigger something. The trap: bigger B B B faster , so the period gets smaller . Fix: derive it — one cycle needs
B Δ x = 2 π ⇒ T = 2 π / ∣ B ∣ B\Delta x = 2\pi \Rightarrow T=2\pi/|B| B Δ x = 2 π ⇒ T = 2 π /∣ B ∣ B B B bottom .
Common mistake "Phase shift is− C -C − C Why it feels right: + C +C + C C C C B = 1 B=1 B = 1 The trap: when B ≠ 1 B\neq1 B = 1 B B B Fix: shift = − C / B =-C/B = − C / B − C -C − C
Common mistake "Shift direction:+ C +C + C Why it feels right: we read "+" as "forward/right." The trap: a change inside the function
is counter-intuitive: f ( x + c ) f(x+c) f ( x + c ) left . Fix: rewrite as B ( x − h ) B(x-h) B ( x − h ) h h h is the true (rightward-positive) shift.
A A A Fix: amplitude = ∣ A ∣ ≥ 0 =|A|\ge0 = ∣ A ∣ ≥ 0 reflection , not a smaller height.
Recall Explain simply (click to reveal)Imagine a jump rope you swing. A A A A A A B B B C C C D D D
"A mplitude B elow-for-period, C -over-B slides, D own/up midline."
A = height, B goes below in 2 π / B 2\pi/B 2 π / B C/B = slide, D = up/down level.
Or: "Big B, Brief period" (bigger B → shorter period).
What is the amplitude of y = A sin ( B x + C ) + D y=A\sin(Bx+C)+D y = A sin ( B x + C ) + D ∣ A ∣ |A| ∣ A ∣ (a non-negative distance; the sign only reflects).
Derive the period of A sin ( B x + C ) + D A\sin(Bx+C)+D A sin ( B x + C ) + D One cycle needs input change
2 π 2\pi 2 π :
B Δ x = 2 π ⇒ T = 2 π / ∣ B ∣ B\,\Delta x=2\pi\Rightarrow T=2\pi/|B| B Δ x = 2 π ⇒ T = 2 π /∣ B ∣ .
Why is B B B Larger
B B B makes the input reach
2 π 2\pi 2 π over a smaller
x x x -range, so the wave repeats faster → shorter period.
What is the phase shift of A sin ( B x + C ) + D A\sin(Bx+C)+D A sin ( B x + C ) + D − C / B -C/B − C / B ; positive = right, negative = left.
Why must you factor B B B The shift is measured in
x x x , but
C C C is added to
B x Bx B x ; factoring gives
B ( x + C / B ) B(x+C/B) B ( x + C / B ) , revealing the true horizontal slide
C / B C/B C / B .
What does D D D Lifts the graph; midline is
y = D y=D y = D , with max
D + ∣ A ∣ D+|A| D + ∣ A ∣ and min
D − ∣ A ∣ D-|A| D − ∣ A ∣ .
Given max M M M m m m A A A D D D A = ( M − m ) / 2 A=(M-m)/2 A = ( M − m ) /2 ,
D = ( M + m ) / 2 D=(M+m)/2 D = ( M + m ) /2 .
What does a negative A A A Reflects it across the midline (flips it upside down); amplitude stays
∣ A ∣ |A| ∣ A ∣ .
For y = 3 sin ( 2 x − π / 2 ) + 1 y=3\sin(2x-\pi/2)+1 y = 3 sin ( 2 x − π /2 ) + 1 Period
π \pi π , shift
π / 4 \pi/4 π /4 right, midline
y = 1 y=1 y = 1 .
Does f ( x + c ) f(x+c) f ( x + c ) c > 0 c>0 c > 0 Left — inside changes act opposite to intuition.
Unit circle and sine definition — where sin \sin sin [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 2 π 2\pi 2 π Function transformations — general rules for a f ( b ( x − h ) ) + k af(b(x-h))+k a f ( b ( x − h )) + k cos as shifted sin — cos x = sin ( x + π / 2 ) \cos x=\sin(x+\pi/2) cos x = sin ( x + π /2 ) Simple harmonic motion — physics uses x = A sin ( ω t + ϕ ) x=A\sin(\omega t+\phi) x = A sin ( ω t + ϕ ) Solving trig equations — knowing the transformed graph tells you how many solutions exist.
Intuition Hinglish mein samjhoDekho, koi bhi "lehardaar" sine wave asal mein sirf basic sin x \sin x sin x A , B , C , D A, B, C, D A , B , C , D A A A oonchi hai —
yeh amplitude hai, aur hamesha ∣ A ∣ |A| ∣ A ∣ A A A
B B B period yaani wave kitni jaldi repeat hoti hai. Formula T = 2 π / ∣ B ∣ T=2\pi/|B| T = 2 π /∣ B ∣ B x Bx B x 2 π 2\pi 2 π B Δ x = 2 π B\,\Delta x=2\pi B Δ x = 2 π T = 2 π / ∣ B ∣ T=2\pi/|B| T = 2 π /∣ B ∣ bada B B B (wave tez
repeat karti hai), isiliye B B B
C C C phase shift aata hai, par yahaan students phaste hain. Pehle B B B B x + C = B ( x + C / B ) Bx+C = B(x + C/B) B x + C = B ( x + C / B ) = − C / B =-C/B = − C / B + + + ulta (left) shift karta hai. D D D y = D y=D y = D = D + ∣ A ∣ =D+|A| = D + ∣ A ∣ = D − ∣ A ∣ =D-|A| = D − ∣ A ∣
Yeh matter isliye karta hai kyunki physics ki SHM, sound, light, AC current — sab isi
A sin ( B x + C ) + D A\sin(Bx+C)+D A sin ( B x + C ) + D
