AP EAPCET (Engineering) · Mathematics (JEE & NDA)
Trigonometry
Trigonometric identities and equations, inverse trigonometric functions, and heights and distances applications.
Seven concepts. JEE trigonometry is identities you rewrite until a standard equation or a right triangle appears — not a catalogue of forms to memorise in isolation.
- AP EAPCET (Engineering)
- Medium level
- 7 concepts
- 5 practice questions
1Pythagorean identities
Every simplification that looks messy starts from three relations that live on the unit circle: \sin^2\theta+\cos^2\theta=1, 1+\tan^2\theta=\sec^2\theta and 1+\cot^2\theta=\csc^2\theta. The first is the statement that a point (\cos\theta,\sin\theta) lies on x^2+y^2=1; the other two are that same statement divided through by \cos^2\theta or \sin^2\theta.
When an identity looks stuck, rewrite every ratio in sin and cos and cancel against one of these three. That is almost always cheaper than inventing a new trick for the form in front of you.
Figure. On the unit circle a point is (\cos\theta,\sin\theta), so \cos^2\theta+\sin^2\theta=1 is the circle equation itself.
How they follow
- Start from the circle\sin^2\theta+\cos^2\theta=1 is the radius-1 condition in trig clothing.
- Divide by \cos^2\thetaWherever \cos\theta\neq 0, you get 1+\tan^2\theta=\sec^2\theta.
- Divide by \sin^2\thetaWherever \sin\theta\neq 0, you get 1+\cot^2\theta=\csc^2\theta.
From the circle to the secant form
Derive 1+\tan^2\theta=\sec^2\theta from \sin^2\theta+\cos^2\theta=1, for \cos\theta\neq 0.
- \sin^2\theta+\cos^2\theta=1given
- divide through by \cos^2\theta\dfrac{\sin^2\theta}{\cos^2\theta}+1=\dfrac{1}{\cos^2\theta}
- rewrite the ratios\tan^2\theta+1=\sec^2\theta
Pro tip. Express everything in \sin and \cos when an identity looks messy; most simplifications collapse once written this way.
If \cos\theta=0, which of the three Pythagorean identities is unavailable?
- \sin^2\theta+\cos^2\theta=1
- 1+\tan^2\theta=\sec^2\theta
- 1+\cot^2\theta=\csc^2\theta
Dividing by \cos^2\theta is how 1+\tan^2\theta=\sec^2\theta is obtained, and that step is illegal when \cos\theta=0. The first identity still holds (\sin^2\theta=1), and the cot/csc form divides by \sin^2\theta instead.
2Compound and double-angle formulas
Adding angles is not a product of ratios: \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B and \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B. Set A=B=\theta and the double-angle forms fall out: \sin 2\theta=2\sin\theta\cos\theta and \cos 2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta.
The three writings of \cos 2\theta are the same identity in different clothes. Pick the one that matches the power you already have: 1-2\sin^2\theta when the expression is in \sin, 2\cos^2\theta-1 when it is in \cos.
Angle-addition formulas are algebraic rewrites in \sin and \cos; expand and collect on the page.
Choosing a form
- Name the target angleIf the problem asks for 2\theta or A+B, reach for a compound or double-angle identity rather than expanding from scratch.
- Match the powerFor \cos 2\theta, use 1-2\sin^2\theta beside a \sin^2 expression and 2\cos^2\theta-1 beside a \cos^2 one.
- Fall back to \sin and \cosIf the form still looks foreign, expand every compound angle fully into \sin and \cos and cancel.
| Form | Identity |
|---|---|
| \sin(A\pm B) | \sin A\cos B\pm\cos A\sin B |
| \cos(A\pm B) | \cos A\cos B\mp\sin A\sin B |
| \sin 2\theta | 2\sin\theta\cos\theta |
| \cos 2\theta | 1-2\sin^2\theta or 2\cos^2\theta-1 |
To rewrite 1-2\sin^2 3x as a single cosine, the matching double-angle form is
- \cos 6x
- \cos 3x
- \sin 6x
\cos 2\theta=1-2\sin^2\theta with \theta=3x gives \cos 6x. Using \theta=x would leave the argument at 2x, and the sine double-angle form is a product, not a 1-2\sin^2 shape.
3The amplitude form a\cos\theta+b\sin\theta
A linear combination a\cos\theta+b\sin\theta is one cosine with a phase shift: R\cos(\theta-\phi) where R=\sqrt{a^2+b^2}, \cos\phi=a/R and \sin\phi=b/R. Once it is in that shape, its range is exactly [-R, R] — so maximum and minimum questions die in one line.
The same rewrite solves equations of the form a\cos\theta+b\sin\theta=c: divide by R and read a cosine equal to c/R, provided |c|\le R.
Figure. 3\cos\theta+4\sin\theta is a single wave of amplitude R=\sqrt{3^2+4^2}=5, so the range is exactly [-5,5].
Packing into one cosine
- Compute RR=\sqrt{a^2+b^2}. That single number is both the amplitude and the half-width of the range.
- Read the phase\cos\phi=a/R and \sin\phi=b/R, so \tan\phi=b/a with the quadrant fixed by the signs of a and b.
- Rewrite and finisha\cos\theta+b\sin\theta=R\cos(\theta-\phi). Max/min are \pm R; an equation becomes \cos(\theta-\phi)=c/R.
Packing 3\cos\theta+4\sin\theta
Rewrite 3\cos\theta+4\sin\theta as R\cos(\theta-\phi) and state its range.
- R=\sqrt{3^2+4^2}5
- \cos\phi=3/5, \sin\phi=4/5\tan\phi=4/3 (first quadrant)
- 3\cos\theta+4\sin\theta5\cos(\theta-\phi)
- range of \cos is [-1,1]range [-5,5]
Pro tip. For a\cos\theta+b\sin\theta, use the amplitude form R\cos(\theta-\phi) with R=\sqrt{a^2+b^2}; its range is [-R,R], answering max/min instantly.
The maximum value of 5\cos\theta-12\sin\theta is
- 5
- 13
- 17
R=\sqrt{5^2+(-12)^2}=\sqrt{25+144}=13, so the combination reaches +13 and -13. Adding the coefficients (17) or reading only the cosine coefficient (5) both skip the amplitude.
4General solutions of trigonometric equations
A trigonometric equation is solved in two moves: reduce it to a statement about one ratio, then write the general angle that produces that value. The three stock forms are \sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha, \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha, and \tan\theta=\tan\alpha\Rightarrow\theta=n\pi+\alpha, for integer n.
When the equation is quadratic in a ratio — for example in \sin\theta — factor or use the quadratic formula first, then write a general solution for each root separately. Do not merge the two families into one formula.
Wrap a principal value by the correct integer family (2n\pi\pm\alpha or n\pi+\alpha) — a periodicity checklist.
Solving for the angle
- Reduce to one ratioFactor, substitute t=\tan\theta, or use an identity until you have \sin, \cos or \tan equal to a constant.
- Read the principal valueName an \alpha in the usual range with that sine, cosine or tangent.
- Write the familyApply the matching general-solution formula, once per root if there are several.
| Equation | General solution |
|---|---|
| \sin\theta=\sin\alpha | \theta=n\pi+(-1)^n\alpha |
| \cos\theta=\cos\alpha | \theta=2n\pi\pm\alpha |
| \tan\theta=\tan\alpha | \theta=n\pi+\alpha |
Solving a trigonometric equation
Find the general solution of 2\sin^2\theta-\sin\theta-1=0.
- let s=\sin\theta; 2s^2-s-1=0(2s+1)(s-1)=0
- 2s+1=0 or s-1=0s=-1/2 or s=1
- \sin\theta=1\theta=\dfrac{\pi}{2}+2n\pi
- \sin\theta=-\dfrac{1}{2}\theta=n\pi+(-1)^n\left(-\dfrac{\pi}{6}\right)
Pro tip. Treat the equation as a quadratic in \sin\theta; solve for the ratio first, then write each general solution separately.
If \sin\theta=-1/2, one correct general solution is
- \theta=n\pi+(-1)^n(-\pi/6)
- \theta=2n\pi+\pi/6
- \theta=n\pi+\pi/6
The sine general solution carries the (-1)^n flip; dropping it (or switching to the cosine \pm form) misses every other solution in the family. \theta=2n\pi+\pi/6 is a cosine-shaped family for a positive sine value, not for -1/2.
5Inverse trigonometric ranges and identities
An inverse trig value is an angle in a fixed principal range, not any angle whose sine happens to match. By convention \sin^{-1}x\in[-\pi/2,\pi/2], \cos^{-1}x\in[0,\pi] and \tan^{-1}x\in(-\pi/2,\pi/2). Outside those ranges the symbol is not the principal value the exam means.
Two identities earn their keep constantly: \sin^{-1}x+\cos^{-1}x=\pi/2 on [-1,1], and \tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy} when xy<1. The xy<1 guard is the whole trap — when xy>1 the sum jumps by \pi into another branch.
Principal values live in fixed intervals ([-\pi/2,\pi/2] for arcsin); membership in that interval is the content.
Reading an inverse
- Name the rangeBefore simplifying, recall which interval the inverse is defined to land in.
- Apply a named identity\sin^{-1}x+\cos^{-1}x=\pi/2 cancels a pair; the two-argument \tan^{-1} formula needs xy<1 checked first.
- Reject out-of-range answersIf a calculation produces an angle outside the principal range, it is not the value of the inverse.
| Function | Domain | Principal range |
|---|---|---|
| \sin^{-1}x | [-1,1] | [-\pi/2,\pi/2] |
| \cos^{-1}x | [-1,1] | [0,\pi] |
| \tan^{-1}x | \mathbb{R} | (-\pi/2,\pi/2) |
The value of \sin^{-1}(1/2)+\cos^{-1}(1/2) is
- \pi/2
- \pi/3
- \pi/6
\sin^{-1}x+\cos^{-1}x=\pi/2 for every x in [-1,1], including x=1/2. Computing \sin^{-1}(1/2)=\pi/6 and stopping there, or adding \pi/6+\pi/3 and somehow dropping a term, are how \pi/6 and \pi/3 appear as distractors.
6Sine rule and cosine rule
In any triangle, the sine rule \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R ties each side to the sine of the opposite angle and to the circumradius R. The cosine rule \cos A=\dfrac{b^2+c^2-a^2}{2bc} (and its cyclic partners) recovers an angle from three sides, or a side from the included angle.
Use the sine rule when you know a side-angle opposite pair and one more piece; use the cosine rule when you know three sides, or two sides and the included angle. Ambiguous-case SSA is the sine-rule trap — two triangles can share the same data.
Figure. Side a sits opposite angle A, side b opposite B, side c opposite C. The sine rule pairs each side with the sine of its opposite angle; the cosine rule rebuilds one angle from the three sides.
Which rule
- Inventory the dataList the known sides and angles and mark any opposite pair.
- Pick sine or cosineOpposite pair plus one more → sine rule. Three sides, or SAS → cosine rule.
- Watch SSAWith two sides and a non-included angle, check whether a second triangle fits before committing to one answer.
| Rule | Statement | Reach for it when |
|---|---|---|
| Sine | a/\sin A=b/\sin B=c/\sin C=2R | An opposite side-angle pair is known |
| Cosine | \cos A=(b^2+c^2-a^2)/(2bc) | Three sides, or two sides and the included angle |
Given sides a,b,c and no angles, the direct route to angle A is
- The sine rule, reading \sin A from a
- The cosine rule, reading \cos A from a,b,c
- The amplitude form R\cos(A-\phi)
Three sides make the cosine rule immediate. The sine rule would need an angle first (or produce the ambiguous SSA situation if you somehow invented one), and the amplitude form belongs to a\cos\theta+b\sin\theta, not to a triangle.
7Heights and distances
Elevation and depression problems are two right triangles that share a vertical side — the tower, the building, the cliff. Write one \tan equation per observation point, then eliminate the shared base distance between them.
The geometry is ordinary: if the farther angle of elevation is \alpha and the nearer is \beta, with horizontal separation d between the points, the height satisfies h(\cot\alpha-\cot\beta)=d. Deriving that once from the two \tan equations is better than memorising it.
Figure. Angles drawn to scale for aspect 1.6: the farther sight line rises at 30° and the nearer at 60°. The 40 m bracket is the walk between the observation points; h is what the two tan equations solve for.
Two tans, one height
- Sketch the shared uprightDraw the tower once; mark both observation points on the ground and the two elevation angles.
- One equation eachNearer point: \tan\beta=h/x. Farther point: \tan\alpha=h/(x+d).
- Eliminate xSolve each for the base and subtract, or equate after clearing h, until only h and the given d remain.
Height and distance
From a point the angle of elevation of a tower top is 30^\circ; moving 40\text{ m} closer it becomes 60^\circ. Find the tower height.
- \tan 60^\circ=h/xx=h/\sqrt{3}
- \tan 30^\circ=h/(x+40)x+40=h\sqrt{3}
- h\sqrt{3}-h/\sqrt{3}=402h/\sqrt{3}=40
- h=40\cdot\sqrt{3}/2h=20\sqrt{3}\approx 34.6\text{ m}
Pro tip. In heights-and-distances problems, look for a common side shared by two right triangles and eliminate it between the two \tan equations.
In the worked example, the distance from the nearer point to the foot of the tower is
- 20\text{ m}
- 40\text{ m}
- 20\sqrt{3}\text{ m}
From x=h/\sqrt{3} and h=20\sqrt{3}, x=20\text{ m}. The 40\text{ m} is the walk between the two observation points, not the nearer base; 20\sqrt{3}\text{ m} is the height itself.
Notes
- Fundamental identities: \sin^2\theta+\cos^2\theta=1, 1+\tan^2\theta=\sec^2\theta and 1+\cot^2\theta=\csc^2\theta relate the ratios. These follow from the unit circle and underpin all simplifications.
- Compound and multiple angles: \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B and \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B; doubling gives \sin2\theta=2\sin\theta\cos\theta and \cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta.
- Trigonometric equations: The general solutions are \sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha, \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha and \tan\theta=\tan\alpha\Rightarrow\theta=n\pi+\alpha, for integer n.
- Inverse trigonometric functions: These are defined on restricted ranges, e.g. \sin^{-1}x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]. Useful identities include \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2} and \tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy} (for xy<1).
- Properties of triangles: The sine rule \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R and cosine rule \cos A=\dfrac{b^2+c^2-a^2}{2bc} solve triangles; heights and distances problems apply these with angles of elevation and depression.
Formulas
- Pythagorean: \sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta
- Compound angle: \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B
- Double angle: \sin2\theta=2\sin\theta\cos\theta,\quad \cos2\theta=1-2\sin^2\theta
- General solutions: \sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha
- Inverse identity: \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}
- Sine / cosine rule: \dfrac{a}{\sin A}=2R,\quad \cos A=\dfrac{b^2+c^2-a^2}{2bc}
Exam traps & shortcuts
- Express everything in \sin and \cos when an identity looks messy; most simplifications collapse once written this way.
- For a\cos\theta+b\sin\theta, use the amplitude form R\cos(\theta-\phi) with R=\sqrt{a^2+b^2}; its range is [-R,R], answering max/min instantly.
- In heights-and-distances problems, look for a common side shared by two right triangles and eliminate it between the two \tan equations.
Reference tables
The forms that unlock most JEE trig simplifications, collected once so the concept cards can stay on method.
| Family | Keep ready |
|---|---|
| Pythagorean | \sin^2\theta+\cos^2\theta=1, 1+\tan^2\theta=\sec^2\theta |
| Compound / double | \sin(A\pm B), \cos(A\pm B), \sin 2\theta, \cos 2\theta=1-2\sin^2\theta |
| Amplitude | a\cos\theta+b\sin\theta=R\cos(\theta-\phi), R=\sqrt{a^2+b^2} |
| General solution | \sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha |
| Inverse | \sin^{-1}x+\cos^{-1}x=\pi/2; \tan^{-1} sum needs xy<1 |
| Triangle | a/\sin A=2R; \cos A=(b^2+c^2-a^2)/(2bc) |
Recap
Read only this the night before.
- Sin / cos first
- Messy identity → write every ratio in sin and cos, then cancel against a Pythagorean form.
- Amplitude
- a cos θ + b sin θ has range exactly [−R, R] with R = √(a² + b²). Max/min is R, not a + b.
- General solution
- Solve for the ratio first; write one family per root. Sin carries (−1)ⁿ; cos carries ±; tan is period π.
- Inverse ranges
- sin⁻¹ lands in [−π/2, π/2], cos⁻¹ in [0, π]. tan⁻¹x + tan⁻¹y needs xy < 1 before the usual sum formula.
- Heights
- Two observation points → two tan equations → eliminate the shared base. The walk between points is not the nearer distance to the tower.
Practise Trigonometry
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