Intuition The big pictureSine, cosine and tangent are periodic and many-to-one . So an equation like sin θ = 1 2 \sin\theta = \tfrac12 sin θ = 2 1 infinitely many answers. A "general solution" is a single formula that packs ALL of them using an integer n n n n n n
WHY it matters: without a general solution you'd forever be guessing angles. The formula is a machine: turn the crank (n = 0 , ± 1 , ± 2 , … n = 0, \pm1, \pm2,\dots n = 0 , ± 1 , ± 2 , …
A function is periodic with period T T T f ( θ + T ) = f ( θ ) f(\theta + T) = f(\theta) f ( θ + T ) = f ( θ ) θ \theta θ
sin \sin sin cos \cos cos = = 2 π = = ==2\pi== == 2 π == 360 ∘ 360^\circ 36 0 ∘ tan \tan tan = = π = = ==\pi== == π == 180 ∘ 180^\circ 18 0 ∘
HOW this creates solutions: if θ 0 \theta_0 θ 0
But sine and cosine each hit a value at two places per cycle (except peaks). So we need two families of angles for them, one family for tan.
Intuition WHY two families collapse into± \pm ± cos \cos cos even function: cos ( − θ ) = cos θ \cos(-\theta) = \cos\theta cos ( − θ ) = cos θ α \alpha α − α -\alpha − α y y y
Let α = cos − 1 c \alpha = \cos^{-1}c α = cos − 1 c principal angle. The two solutions in one turn are θ = α \theta = \alpha θ = α θ = − α \theta = -\alpha θ = − α 2 π − α 2\pi - \alpha 2 π − α 2 π 2\pi 2 π
Why the ± \pm ± because of even symmetry. Why 2 n π 2n\pi 2 nπ because that's the period.
( − 1 ) n (-1)^n ( − 1 ) n sin \sin sin odd and symmetric about θ = π 2 \theta = \tfrac{\pi}{2} θ = 2 π α \alpha α π − α \pi - \alpha π − α
n n n → \to → α \alpha α n n n → \to → π − α \pi-\alpha π − α
The trick ( − 1 ) n (-1)^n ( − 1 ) n α \alpha α n n n n π n\pi nπ
Let α = sin − 1 s \alpha = \sin^{-1}s α = sin − 1 s
Check it works: n = 0 ⇒ α n=0 \Rightarrow \alpha n = 0 ⇒ α n = 1 ⇒ π − α n=1 \Rightarrow \pi - \alpha n = 1 ⇒ π − α n = 2 ⇒ 2 π + α n=2 \Rightarrow 2\pi + \alpha n = 2 ⇒ 2 π + α
Intuition WHY only ONE family (no± \pm ± tan \tan tan π \pi π exactly once . So one base angle plus multiples of π \pi π
Let α = tan − 1 t \alpha = \tan^{-1}t α = tan − 1 t
Rearrange to sin / cos / tan ( angle ) = number \sin/\cos/\tan(\text{angle}) = \text{number} sin / cos / tan ( angle ) = number
Find the principal value α \alpha α
Write the general solution formula.
Substitute n = … , − 1 , 0 , 1 , 2 , … n = \dots,-1,0,1,2,\dots n = … , − 1 , 0 , 1 , 2 , … keep only the θ \theta θ
If the equation has a multiple angle (e.g. cos 2 θ \cos 2\theta cos 2 θ
Worked example Example 2 — sine using( − 1 ) n (-1)^n ( − 1 ) n sin θ = 3 2 \sin\theta = \tfrac{\sqrt3}{2} sin θ = 2 3 0 ≤ θ ≤ 2 π 0 \le \theta \le 2\pi 0 ≤ θ ≤ 2 π
α = sin − 1 3 2 = π 3 \alpha = \sin^{-1}\!\tfrac{\sqrt3}{2} = \tfrac{\pi}{3} α = sin − 1 2 3 = 3 π General: θ = n π + ( − 1 ) n π 3 \theta = n\pi + (-1)^n\tfrac{\pi}{3} θ = nπ + ( − 1 ) n 3 π
n = 0 n=0 n = 0 θ = π 3 \theta = \tfrac{\pi}{3} θ = 3 π Why keep? in range.n = 1 n=1 n = 1 θ = π − π 3 = 2 π 3 \theta = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3} θ = π − 3 π = 3 2 π Why the minus? ( − 1 ) 1 = − 1 (-1)^1=-1 ( − 1 ) 1 = − 1 n = 2 n=2 n = 2 θ = 2 π + π 3 \theta = 2\pi + \tfrac{\pi}{3} θ = 2 π + 3 π Answers: π 3 , 2 π 3 \tfrac{\pi}{3},\ \tfrac{2\pi}{3} 3 π , 3 2 π
Worked example Example 3 — tangentSolve tan θ = − 1 \tan\theta = -1 tan θ = − 1 0 ∘ ≤ θ < 360 ∘ 0^\circ \le \theta < 360^\circ 0 ∘ ≤ θ < 36 0 ∘
α = tan − 1 ( − 1 ) = − 45 ∘ \alpha = \tan^{-1}(-1) = -45^\circ α = tan − 1 ( − 1 ) = − 4 5 ∘ Why negative? tan is odd; calculator gives − 45 ∘ -45^\circ − 4 5 ∘ General: θ = 180 ∘ n − 45 ∘ \theta = 180^\circ n - 45^\circ θ = 18 0 ∘ n − 4 5 ∘
n = 1 n=1 n = 1 135 ∘ 135^\circ 13 5 ∘ n = 2 n=2 n = 2 315 ∘ 315^\circ 31 5 ∘ n = 0 n=0 n = 0 − 45 ∘ -45^\circ − 4 5 ∘ n = 3 n=3 n = 3 495 ∘ 495^\circ 49 5 ∘ Answers: 135 ∘ , 315 ∘ 135^\circ,\ 315^\circ 13 5 ∘ , 31 5 ∘
Recall Before reading the answer, predict: How many solutions doessin θ = 0.4 \sin\theta = 0.4 sin θ = 0.4 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π 0 ≤ θ < 4 π 0\le\theta<4\pi 0 ≤ θ < 4 π 2 2 2 2 π 2\pi 2 π 2 2 2 4 4 4 n n n
Common mistake Only giving the calculator answerWrong idea: "cos θ = 1 2 ⇒ θ = 60 ∘ \cos\theta=\tfrac12 \Rightarrow \theta = 60^\circ cos θ = 2 1 ⇒ θ = 6 0 ∘ Why it feels right: the calculator gives exactly one value, so it looks complete.
Fix: the calculator gives only the principal value. Use the general solution to get the whole family, then filter by range. You'd have missed 300 ∘ 300^\circ 30 0 ∘
Common mistake Forgetting to widen the range for multiple anglesWrong idea: solve cos 2 θ = 3 2 \cos 2\theta = \tfrac{\sqrt3}{2} cos 2 θ = 2 3 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π 2 θ 2\theta 2 θ Why it feels right: the stated range is 2 π 2\pi 2 π Fix: the substitution ϕ = 2 θ \phi=2\theta ϕ = 2 θ doubles the interval to 0 ≤ ϕ < 4 π 0\le\phi<4\pi 0 ≤ ϕ < 4 π
Common mistake Sign confusion with( − 1 ) n (-1)^n ( − 1 ) n Wrong idea: applying ± \pm ± Why it feels right: both look symmetric.
Fix: sine's two solutions are α \alpha α π − α \pi-\alpha π − α not ± α \pm\alpha ± α n π + ( − 1 ) n α n\pi+(-1)^n\alpha nπ + ( − 1 ) n α 2 n π ± α 2n\pi\pm\alpha 2 nπ ± α
Recall Explain to a 12-year-oldImagine a Ferris wheel that keeps spinning forever. If I ask "when is your seat exactly 3 metres high?", there isn't just one time — it happens every loop, twice each loop (once going up, once coming down). The general solution is like saying: "at these two clock positions, and then again every full turn after." The little number n n n which loop you're on. If you only want answers before lunch, you keep only the loops that finish before lunch.
Mnemonic Remember the three formulas"Cos Plus-minus, Sin Signs-alternate, Tan Alone."
C os → 2 n π ± α 2n\pi \pm \alpha 2 nπ ± α S in → n π + ( − 1 ) n α n\pi + (-1)^n\alpha nπ + ( − 1 ) n α T an → n π + α n\pi + \alpha nπ + α π \pi π
General solution of cos θ = cos α \cos\theta=\cos\alpha cos θ = cos α θ = 2 n π ± α , n ∈ Z \theta = 2n\pi \pm \alpha,\ n\in\mathbb Z θ = 2 nπ ± α , n ∈ Z General solution of sin θ = sin α \sin\theta=\sin\alpha sin θ = sin α θ = n π + ( − 1 ) n α , n ∈ Z \theta = n\pi + (-1)^n\alpha,\ n\in\mathbb Z θ = nπ + ( − 1 ) n α , n ∈ Z General solution of tan θ = tan α \tan\theta=\tan\alpha tan θ = tan α θ = n π + α , n ∈ Z \theta = n\pi + \alpha,\ n\in\mathbb Z θ = nπ + α , n ∈ Z Why does cosine's general solution use ± \pm ± Because
cos \cos cos is even,
cos ( − α ) = cos α \cos(-\alpha)=\cos\alpha cos ( − α ) = cos α , giving two symmetric roots per turn.
Why does tangent need only one family? Its period is
π \pi π and it hits each value exactly once per period.
When solving cos 2 θ = c \cos 2\theta = c cos 2 θ = c 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π Widen the range to
0 ≤ 2 θ < 4 π 0\le 2\theta<4\pi 0 ≤ 2 θ < 4 π before solving, then divide results by 2.
How many solutions does sin θ = 0.4 \sin\theta = 0.4 sin θ = 0.4 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π Two.
What does the integer n n n Which full period (loop) of the function the solution lies in.
First step in any trig equation? Rearrange to isolate
sin / cos / tan \sin/\cos/\tan sin / cos / tan of the angle equal to a number.
Trig functions periodic and many-to-one
Infinitely many solutions
General solution formula with n
cos: theta = 2n pi ± alpha
sin: theta = n pi + -1^n alpha
Tan hits value once per period
tan: theta = n pi + alpha
Intuition Hinglish mein samjhoDekho, sin, cos aur tan periodic functions hain — matlab woh baar-baar apni values repeat karte hain. Isliye jab tum likhte ho sin θ = 1 2 \sin\theta = \tfrac12 sin θ = 2 1 infinite answers hote hain, kyunki har poore ghoomne (period) ke baad wahi value phir aa jaati hai. General solution ek chhota sa formula hai jo ye saare answers ek saath capture kar leta hai, ek integer n n n n = 0 , 1 , 2 , − 1 … n = 0, 1, 2, -1 \dots n = 0 , 1 , 2 , − 1 …
Teen formule yaad rakhne hain: cosine ke liye θ = 2 n π ± α \theta = 2n\pi \pm \alpha θ = 2 nπ ± α ± \pm ± θ = n π + ( − 1 ) n α \theta = n\pi + (-1)^n\alpha θ = nπ + ( − 1 ) n α α \alpha α π − α \pi-\alpha π − α θ = n π + α \theta = n\pi + \alpha θ = nπ + α π \pi π α \alpha α
Range mein solutions chahiye toh simple: pehle equation ko sin / cos / tan = \sin/\cos/\tan = sin / cos / tan = α \alpha α n n n θ \theta θ cos 2 θ \cos 2\theta cos 2 θ 0 0 0 4 π 4\pi 4 π 2 θ 2\theta 2 θ