Trigonometric Functions

1. Check this before you revise anything

One topic sits in an unusual middle ground for this chapter.

"Formative-only" is CBSE's own phrase, and it means what it says: this will not appear on your board exam paper, even though it technically sits inside the syllabus document. It is meant to be touched on in class discussion, not drilled for marks.

Combined with the fact that this chapter's own text never once writes down the formula, spending revision time memorising $x=n\pi +(-1{)}^n\alpha$ here is solving next chapter's problem, not this one — it belongs to the equation-solving work done later.

2. What this chapter covers

3. Angles: degree measure and radian measure

An angle is a measure of rotation of a ray about its starting point. Rotating anticlockwise gives a positive angle; clockwise gives a negative one.

Degree measure. One complete revolution is $360°$. A degree splits into $60$ minutes ($60^{\prime }$), and a minute into $60$ seconds ($60^{\prime \prime }$).

Radian measure. One radian is the angle subtended at the centre of a circle by an arc whose length equals the radius. For a circle of radius $r$, an arc of length $l$ subtending an angle $\theta$ radians satisfies:

$l=r\theta$

Since the circumference of a unit circle is $2\pi$, one full revolution measures $2\pi$ radians, giving the conversion at the heart of this section:

$\pi \text{ radians}=180°$

$\text{Radian measure}=\frac{\pi }{180}\times \text{Degree measure}\text{Degree measure}=\frac{180}{\pi }\times \text{Radian measure}$

Worked, mirroring the textbook's own Example 3. A central angle of $60°$ intercepts an arc of length $37.4$ cm. Find the radius (use $\pi =22/7$). Convert: $\theta =60°=\pi /3$ radians. From $l=r\theta$: $r=l/\theta =37.4\times 3/\pi =37.4\times 3\times 7/22=35.7$ cm.

Worked, mirroring the textbook's own Example 5. Arcs of the same length in two circles subtend $65°$ and $110°$ at their respective centres. Since $l=r_1{\theta }_1=r_2{\theta }_2$ with ${\theta }_1=13\pi /36$ and ${\theta }_2=22\pi /36$ (both converted to radians), $r_1/r_2={\theta }_2/{\theta }_1=22/13$ — the radii are in the inverse ratio of the angles, since the arc length is held fixed.

4. The unit-circle definition

Take a unit circle (radius $1$) centred at the origin. For a point $P(a,b)$ on the circle where the arc from $(1,0)$ to $P$ has length $x$, define:

$\cos x=a\sin x=b$

Since $O{M}^2+M{P}^2=O{P}^2$ for the right triangle formed by dropping a perpendicular from $P$, this gives the identity every later formula in this chapter is built from:

${\cos }^{2}x+{\sin }^{2}x=1\text{for every real }x$

This is what makes the definition work for any real number, not just angles between $0°$ and $90°$ — $x$ is simply an arc length (or, equivalently, a radian measure), and arc length makes sense whatever direction or however many times you go around the circle.

Quadrantal angles. One full revolution is $2\pi$, so $\angle AOB=\pi /2$, $\angle AOC=\pi$, $\angle AOD=3\pi /2$, with coordinates $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$ respectively:

Periodicity falls straight out of the geometry. Going one full revolution around the circle returns to the same point, so for any integer $n$:

$\sin (2n\pi +x)=\sin x\cos (2n\pi +x)=\cos x$

The remaining four functions are defined from these two:

$\mathrm{cosec}x=\frac{1}{\sin x},x\ne n\pi \sec x=\frac{1}{\cos x},x\ne (2n+1)\frac{\pi }{2}$ $\tan x=\frac{\sin x}{\cos x},x\ne (2n+1)\frac{\pi }{2}\cot x=\frac{\cos x}{\sin x},x\ne n\pi$

Dividing ${\cos }^{2}x+{\sin }^{2}x=1$ through by ${\cos }^{2}x$ and by ${\sin }^{2}x$ gives the other two Pythagorean identities:

$1+{\tan }^{2}x={\sec }^{2}x1+{\cot }^{2}x={\mathrm{cosec}}^{2}x$

5. Signs, domain, and range

Since every point on the unit circle has $-1\le a\le 1$ and $-1\le b\le 1$:

$-1\le \sin x\le 1-1\le \cos x\le 1\text{for every real }x$

Worked, mirroring the textbook's own Example 6. $\cos x=-3/5$, $x$ in the third quadrant. Find the other five. From ${\sin }^{2}x=1-{\cos }^{2}x=1-9/25=16/25$, so $\sin x=\pm 4/5$ — and since $x$ is in the third quadrant, $\sin x$ is negative: $\sin x=-4/5$.

That fixes every remaining ratio: $\tan x=\sin x/\cos x=4/3$, $\cot x=3/4$, $\sec x=-5/3$, $\mathrm{cosec}x=-5/4$.

The quadrant check is the step that actually matters here — the algebra only ever gives a $\pm$, and skipping the quadrant means guessing the sign instead of determining it. Worked, mirroring the textbook's own Example 7, this cuts the other way too: given $\cot x=-5/12$ with $x$ in the second quadrant, $\sec x$ must be negative there (cosine is negative in QII), forcing $\sec x=-13/5$ even though ${\sec }^{2}x=169/25$ algebraically allows either sign.

6. Allied angles

Six angle-transformations, all provable directly from the unit-circle definition, let you reduce any trigonometric value to one involving an angle between $0$ and $\pi /2$:

Read the pattern instead of memorising six rows separately. Swapping $\pi /2\pm x$ swaps sine and cosine (a "co-function" flip); $\pi \pm x$ and $2\pi -x$ keep the same function but adjust the sign to match whichever quadrant the transformed angle actually lands in. $\tan$, $\cot$, $\sec$, $\mathrm{cosec}$ follow the same six transformations through $\tan x=\sin x/\cos x$ — the book does not tabulate them separately, and neither does this chapter.

Worked, mirroring the textbook's own Example 8 and Example 9. Both use periodicity, not the allied-angle table, to collapse a large angle: $\sin (31\pi /3)=\sin (10\pi +\pi /3)=\sin (\pi /3)=\sqrt{3}/2$, since $10\pi$ is five full revolutions. Similarly $\cos (-1710°)=\cos (-1710°+5\times 360°)=\cos (90°)=0$ — always add or subtract a whole multiple of $360°$ (or $2\pi$), never an arbitrary convenient number.

7. Sum and difference formulas

$\cos (x+y)=\cos x\cos y-\sin x\sin y\cos (x-y)=\cos x\cos y+\sin x\sin y$ $\sin (x+y)=\sin x\cos y+\cos x\sin y\sin (x-y)=\sin x\cos y-\cos x\sin y$

The $\cos (x-y)$ form is proved first, directly from the unit circle using the distance formula; every other identity in this section — including both sine formulas — is then derived from it. Dividing the sine and cosine sum formulas gives the tangent versions, valid whenever none of $x$, $y$, $x\pm y$ is an odd multiple of $\pi /2$:

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

Worked, mirroring the textbook's own Example 11. Find $\sin 15°$. Write $15°=45°-30°$: $\sin 15°=\sin 45°\cos 30°-\cos 45°\sin 30°=\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Worked, mirroring the textbook's own Example 18. $\sin x=3/5$, $\cos y=-12/13$, both $x,y$ in the second quadrant; find $\sin (x+y)$. In QII, $\cos x<0$ and $\sin y>0$, so $\cos x=-4/5$ and $\sin y=5/13$. Then $\sin (x+y)=\sin x\cos y+\cos x\sin y=\frac{3}{5}\left(\frac{-12}{13}\right)+\frac{-4}{5}\left(\frac{5}{13}\right)=-\frac{56}{65}$.

8. Double and triple angle formulas

Setting $y=x$ in the sum formulas gives the double-angle identities — $\cos 2x$ has three interchangeable forms, useful depending on which function appears in the rest of the problem:

$\sin 2x=2\sin x\cos x\cos 2x={\cos }^{2}x-{\sin }^{2}x=2{\cos }^{2}x-1=1-2{\sin }^{2}x\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$

Setting $x=2x,y=x$ inside $\sin (x+y)$ and $\cos (x+y)$ gives the triple-angle identities:

$\sin 3x=3\sin x-4{\sin }^{3}x\cos 3x=4{\cos }^{3}x-3\cos x$

Worked, mirroring the textbook's own Example 20. Find $\tan (\pi /8)$. Let $y=\tan (\pi /8)$. Since $2\times \pi /8=\pi /4$ and $\tan (\pi /4)=1$, the double-angle formula gives $1=2y/(1-y^2)$, i.e. $y^2+2y-1=0$, so $y=-1\pm \sqrt{2}$. Since $\pi /8$ is in the first quadrant, $y>0$, forcing $\tan (\pi /8)=\sqrt{2}-1$.

Worked, mirroring the textbook's own Example 21. $\tan x=3/4$ with $\pi <x<3\pi /2$ — find $\sin (x/2)$, $\cos (x/2)$, $\tan (x/2)$. Halving the interval, $\pi /2<x/2<3\pi /4$, so $x/2$ is in the second quadrant: $\sin (x/2)>0$, $\cos (x/2)<0$. From ${\sec }^{2}x=1+{\tan }^{2}x=25/16$ and $x$ in QIII (so $\cos x<0$): $\cos x=-4/5$. Then $2{\sin }^2(x/2)=1-\cos x=9/5$ and $2{\cos }^2(x/2)=1+\cos x=1/5$, giving $\sin (x/2)=3/\sqrt{10}$, $\cos (x/2)=-1/\sqrt{10}$, and $\tan (x/2)=-3$.

9. Sum-to-product and product-to-sum formulas

Adding and subtracting the $\cos (x\pm y)$ and $\sin (x\pm y)$ expansions, then substituting $\theta =x+y$, $\varphi =x-y$, converts a sum into a product:

$\cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2}\cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}$ $\sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2}\sin x-\sin y=2\cos \frac{x+y}{2}\sin \frac{x-y}{2}$

Read backwards, the same four identities convert a product into a sum — this is the form Exercise 3.3's harder proofs (ratios of sums like $(\sin 7x-\sin 5x)/(\cos 7x+\cos 5x)$) actually need.

Worked, mirroring the textbook's own Example 16. Prove $(\cos 7x+\cos 5x)/(\sin 7x-\sin 5x)=\cot x$. The numerator becomes $2\cos 6x\cos x$ and the denominator becomes $2\cos 6x\sin x$; dividing both by the shared $2\cos 6x$ factor leaves $\cos x/\sin x=\cot x$.

Summary

• $l=r\theta$ relates arc length, radius, and radian measure; $\pi$ radians $=180°$.
• The unit-circle definition, $\cos x=a$, $\sin x=b$ for the point $(a,b)$, extends sine and cosine to every real number and gives ${\cos }^{2}x+{\sin }^{2}x=1$, the source identity for the whole chapter.
• $\sin x$ and $\cos x$ are periodic with period $2\pi$; ASTC gives the sign of each ratio by quadrant, and the domain/range of all six functions follows directly from $-1\le \sin x,\cos x\le 1$.
• Six allied-angle transformations ($-x$, $\pi /2\pm x$, $\pi \pm x$, $2\pi -x$) reduce any angle to one between $0$ and $\pi /2$; large angles reduce first through periodicity ($2n\pi$), not through the allied-angle table.
• $\cos (x\pm y)$ and $\sin (x\pm y)$ generate the double-angle, triple-angle, and sum-to-product/product-to-sum families by substitution — nothing after Section 7 is an independent result.
• The general solution of trigonometric equations is technically listed in the 2026-27 syllabus but graded formatively only — it does not appear in this chapter's exercises and will not be summatively (board-exam) assessed here.