Punjab Board Class 11 Mathematics

Chapter 10: Trigonometric Identities — Class 11 Maths Notes

Trigonometric identities: the standard LHS-to-RHS proof strategy for every identity family, fully worked. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 10.

Exercise solutions — Chapter 10

Every exercise of Trigonometric Identities solved step by step.

Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4

Mathematics 11 (Punjab Board)

Chapter 10: Trigonometric Identities — Theory

Distance Formula and the Fundamental Law

Do points \(P(x_1, y_1)\) aur \(Q(x_2, y_2)\) ke darmiyan fasla \(d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\) hota hai. Isi formula ko unit circle par apply kar ke trigonometry ka sab se bunyadi qanoon sabit hota hai — Fundamental Law of Trigonometry:

\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta

Saboot mein unit circle par angles \(\alpha\) aur \(\beta\) ke terminal points le kar un ke darmiyan fasle ko do tareeqon se likha jata hai. Ye aik identity tamam compound-angle formulas ki bunyad hai.

Key point: \(\cos(\alpha - \beta)\) ka formula yaad rakho — baqi tamam identities isi se nikalti hain.

Sum and Difference Identities

Fundamental law mein \(\beta\) ki jagah \(-\beta\) rakhne se, aur \(\frac{\pi}{2}\) wali substitutions se, ye identities milti hain:

\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta, \qquad \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}, \qquad \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}

Sine ke formulas mein "sin-cos, cos-sin" ka pattern hai; cosine mein signs ulte hain — \(\cos(\alpha+\beta)\) mein minus aata hai. Tangent ke formulas sine aur cosine ko divide kar ke bante hain.

Key point: \(\sin(\alpha+\beta)\) mein plus, \(\cos(\alpha+\beta)\) mein minus — ye sign ka farq board mein sab se zyada ghalat hota hai.

Allied Angles

Do angles allied kehlate hain agar un ka sum ya difference \(90^\circ\) ka multiple ho. Teen yaad rakhne wale usool: (1) odd multiple of \(90^\circ\) par ratio co-ratio mein badal jata hai (\(\sin \leftrightarrow \cos\), \(\tan \leftrightarrow \cot\), \(\sec \leftrightarrow \csc\)); (2) even multiple par ratio same rehta hai; (3) sign us quadrant se decide hota hai jahan angle ka terminal arm ho (ASTC rule).

\sin(90^\circ - \theta) = \cos\theta, \quad \cos(90^\circ + \theta) = -\sin\theta, \quad \tan(180^\circ + \theta) = \tan\theta

\sin(270^\circ - \theta) = -\cos\theta, \quad \cos(360^\circ - \theta) = \cos\theta, \quad \tan(360^\circ - \theta) = -\tan\theta

Key point: Pehle dekho \(90^\circ\) ka multiple odd hai ya even (ratio change hoga ya nahi), phir quadrant se sign decide karo.

Double Angle Identities

Addition formulas mein \(\beta = \alpha\) rakhne se double angle formulas milte hain:

\sin 2\alpha = 2\sin\alpha\cos\alpha

\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha

\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}

\(\cos 2\alpha\) ke teen forms mein se woh choose karo jo sawal mein di gayi information se match kare — masalan agar sirf \(\sin\alpha\) diya ho to \(1 - 2\sin^2\alpha\) wala form use karo.

Key point: \(\sin 2\alpha\) aur \(\cos 2\alpha\) ko \(\tan\alpha\) mein bhi likha ja sakta hai: \(\sin 2\alpha = \frac{2\tan\alpha}{1+\tan^2\alpha}\), \(\cos 2\alpha = \frac{1-\tan^2\alpha}{1+\tan^2\alpha}\).

Half Angle and Triple Angle Identities

Double angle formulas mein \(\alpha\) ki jagah \(\frac{\alpha}{2}\) rakhne se half angle formulas bante hain:

\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{2}}, \qquad \cos\frac{\alpha}{2} = \pm\sqrt{\frac{1+\cos\alpha}{2}}, \qquad \tan\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}

Sign \(\pm\) ka faisla \(\frac{\alpha}{2}\) ke quadrant se hota hai. Triple angle formulas \(\sin(2\alpha + \alpha)\) expand kar ke milte hain:

\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha, \qquad \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha

Key point: Half angle mein sign bhoolna aam ghalti hai — hamesha quadrant check karo.

Product-to-Sum and Sum-to-Product

Addition/subtraction formulas ko add ya subtract kar ke products sums mein badal jate hain:

2\sin\alpha\cos\beta = \sin(\alpha+\beta) + \sin(\alpha-\beta)

2\cos\alpha\cos\beta = \cos(\alpha+\beta) + \cos(\alpha-\beta), \qquad -2\sin\alpha\sin\beta = \cos(\alpha+\beta) - \cos(\alpha-\beta)

\(\alpha + \beta = P\), \(\alpha - \beta = Q\) rakhne se sum-to-product formulas bante hain:

\sin P + \sin Q = 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}, \qquad \cos P - \cos Q = -2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2}

Key point: \(\cos P - \cos Q\) mein minus sign aata hai — baqi teen formulas mein plus hota hai.

Auxiliary Angle: \(a\sin\theta + b\cos\theta\)

Expression \(a\sin\theta + b\cos\theta\) ko single sine mein likha ja sakta hai. \(a = r\cos\phi\), \(b = r\sin\phi\) rakho:

a\sin\theta + b\cos\theta = r\sin(\theta + \phi), \quad r = \sqrt{a^2+b^2}, \quad \tan\phi = \frac{b}{a}

Agar \(b\) negative ho to form \(r\sin(\theta - \phi)\) ban jati hai. \(\phi\) ka quadrant \(a\) aur \(b\) ke signs se decide hota hai.

Key point: \(r = \sqrt{a^2+b^2}\) hamesha positive hota hai; \(\phi\) ke liye \(\tan\phi = b/a\) use karo.

Conditional Identities for a Triangle

Agar \(\alpha + \beta + \gamma = 180^\circ\) (triangle ke angles) hon to:

\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma

\tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1

Pehli identity \(\tan(\alpha+\beta) = \tan(180^\circ - \gamma) = -\tan\gamma\) se sabit hoti hai.

Key point: Triangle wale sawalon mein pehla qadam hamesha \(\alpha + \beta = 180^\circ - \gamma\) likhna hai.

Mathematics 11 (Punjab Board)

Solved Examples

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 10: Trigonometric Identities.

Example 1: Find the distance between the points (i) \(A(3,8)\), \(B(5,6)\) and (ii) \(P(\cos x, \cos y)\), \(Q(\sin x, \sin y)\)

Distance formula \(d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\) lagayenge. Yehi formula Fundamental Law of Trigonometry ke saboot mein istemal hota hai.

(i)

|AB| = \sqrt{(5-3)^2 + (6-8)^2} = \sqrt{4+4} = 2\sqrt{2}

(ii)

|PQ| = \sqrt{(\cos x - \sin x)^2 + (\cos y - \sin y)^2} = \sqrt{2 - 2\cos x\sin x - 2\cos y\sin y}

Answer: (i) \(2\sqrt{2}\), (ii) \(\sqrt{2 - 2(\cos x\sin x + \cos y\sin y)}\).

Example 2: Find the value of \(\sin\frac{5\pi}{12}\)

\(\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}\) likh kar \(\sin(A+B)\) ka formula lagayenge.

\sin\frac{5\pi}{12} = \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6}

= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}

Answer: \(\sin\frac{5\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}\).

Example 3: Without using tables, write down the values of (i) \(\sin 225^\circ\), (ii) \(\tan 600^\circ\), (iii) \(\cot(-225^\circ)\), (iv) \(\csc(-420^\circ)\)

Pehle \(360^\circ\) ke multiples hata kar angle chota karenge, phir allied angle identities lagayenge.

(i) \(\sin 225^\circ = \sin(180^\circ + 45^\circ) = -\sin 45^\circ = -\frac{1}{\sqrt{2}}\)

(ii) \(\tan 600^\circ = \tan(360^\circ + 240^\circ) = \tan 240^\circ = \tan(180^\circ + 60^\circ) = \tan 60^\circ = \sqrt{3}\)

(iii) \(\cot(-225^\circ) = -\cot 225^\circ = -\cot(180^\circ + 45^\circ) = -\cot 45^\circ = -1\)

(iv) \(\csc(-420^\circ) = -\csc 420^\circ = -\csc(360^\circ + 60^\circ) = -\csc 60^\circ = -\frac{2}{\sqrt{3}}\)

Answer: (i) \(-\frac{1}{\sqrt{2}}\), (ii) \(\sqrt{3}\), (iii) \(-1\), (iv) \(-\frac{2}{\sqrt{3}}\).

Example 4: Simplify \(\frac{\sin(180^\circ-\theta)\cos(360^\circ-\theta)\tan(90^\circ+\theta)}{\sin(90^\circ-\theta)\cos(180^\circ+\theta)\tan(270^\circ-\theta)}\)

Har factor ko allied angle identity se simplify karenge, phir common factors cancel karenge.

\(\sin(180^\circ-\theta) = \sin\theta\); \(\cos(360^\circ-\theta) = \cos\theta\); \(\tan(90^\circ+\theta) = -\cot\theta\); \(\sin(90^\circ-\theta) = \cos\theta\); \(\cos(180^\circ+\theta) = -\cos\theta\); \(\tan(270^\circ-\theta) = \cot\theta\).

= \frac{(\sin\theta)(\cos\theta)(-\cot\theta)}{(\cos\theta)(-\cos\theta)(\cot\theta)} = \frac{\sin\theta\cos\theta}{\cos\theta\cdot\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta

Answer: \(\tan\theta\).

Example 5: Prove that \(\sin(\alpha+\beta)\sin(\alpha-\beta) = \sin^2\alpha - \sin^2\beta = \cos^2\beta - \cos^2\alpha\)

Dono factors ko expand kar ke difference of squares banayenge, phir \(\cos^2 = 1-\sin^2\) substitute karenge.

\text{L.H.S} = (\sin\alpha\cos\beta + \cos\alpha\sin\beta)(\sin\alpha\cos\beta - \cos\alpha\sin\beta) = \sin^2\alpha\cos^2\beta - \cos^2\alpha\sin^2\beta

= \sin^2\alpha(1-\sin^2\beta) - (1-\sin^2\alpha)\sin^2\beta = \sin^2\alpha - \sin^2\beta

= (1-\cos^2\alpha) - (1-\cos^2\beta) = \cos^2\beta - \cos^2\alpha

Answer: The identity is proved.

Example 6: Without using tables, find the values of all trigonometric functions of \(105^\circ\)

\(105^\circ = 60^\circ + 45^\circ\) likh kar sin aur cos ke addition formulas lagayenge; baqi functions un se derive karenge.

\sin 105^\circ = \sin 60^\circ\cos 45^\circ + \cos 60^\circ\sin 45^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}

\cos 105^\circ = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2}-\sqrt{6}}{4}

\tan 105^\circ = \frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}-\sqrt{6}} = -(2+\sqrt{3}), \quad \cot 105^\circ = -(2-\sqrt{3})

\sec 105^\circ = \frac{4}{\sqrt{2}-\sqrt{6}} = -(\sqrt{6}+\sqrt{2}), \quad \csc 105^\circ = \frac{4}{\sqrt{6}+\sqrt{2}} = \sqrt{6}-\sqrt{2}

Answer: \(\sin 105^\circ = \frac{\sqrt{6}+\sqrt{2}}{4}\), \(\cos 105^\circ = \frac{\sqrt{2}-\sqrt{6}}{4}\), \(\tan 105^\circ = -(2+\sqrt{3})\), \(\cot 105^\circ = -(2-\sqrt{3})\), \(\sec 105^\circ = -(\sqrt{6}+\sqrt{2})\), \(\csc 105^\circ = \sqrt{6}-\sqrt{2}\).

Example 7: Prove that \(\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ\)

Numerator aur denominator ko \(\cos 11^\circ\) se divide karenge — \(\tan(45^\circ+11^\circ)\) ka pattern ban jayega.

\text{L.H.S} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - \tan 45^\circ\tan 11^\circ} = \tan(45^\circ + 11^\circ) = \tan 56^\circ = \text{R.H.S}

Answer: The identity is proved.

Example 8: If \(\cos\alpha = -\frac{7}{25}\) and \(\tan\beta = \frac{12}{5}\), where \(\alpha\) is in Quadrant II and \(\beta\) is in Quadrant III, find (i) \(\sin(\alpha+\beta)\), (ii) \(\cos(\alpha+\beta)\). In which quadrant does \(\alpha+\beta\) lie?

Quadrant II mein sin positive aur Quadrant III mein sin aur cos dono negative hote hain. Pehle \(\sin\alpha\), \(\sin\beta\), \(\cos\beta\) nikalenge.

\sin\alpha = \sqrt{1-\frac{49}{625}} = \frac{24}{25}, \qquad \cos\beta = -\frac{5}{13}, \quad \sin\beta = -\frac{12}{13}

\sin(\alpha+\beta) = \left(\frac{24}{25}\right)\left(-\frac{5}{13}\right) + \left(-\frac{7}{25}\right)\left(-\frac{12}{13}\right) = \frac{-120+84}{325} = -\frac{36}{325}

\cos(\alpha+\beta) = \left(-\frac{7}{25}\right)\left(-\frac{5}{13}\right) - \left(\frac{24}{25}\right)\left(-\frac{12}{13}\right) = \frac{35+288}{325} = \frac{323}{325}

Since \(\sin(\alpha+\beta) < 0\) and \(\cos(\alpha+\beta) > 0\), the terminal side of \(\alpha+\beta\) lies in Quadrant IV.

Answer: \(\sin(\alpha+\beta) = -\frac{36}{325}\), \(\cos(\alpha+\beta) = \frac{323}{325}\); \(\alpha+\beta\) lies in Quadrant IV.

Example 9: If \(\alpha, \beta, \gamma\) are the angles of \(\triangle ABC\), prove that (i) \(\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma\) and (ii) \(\tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1\)

\(\alpha+\beta = 180^\circ-\gamma\) se start karenge. Part (i) mein tan ke addition formula, part (ii) mein half angles par wahi trick lagayenge.

(i)

\tan(\alpha+\beta) = \tan(180^\circ-\gamma) = -\tan\gamma \implies \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} = -\tan\gamma

\tan\alpha + \tan\beta = -\tan\gamma + \tan\alpha\tan\beta\tan\gamma \implies \tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma

(ii) \(\frac{\alpha+\beta}{2} = 90^\circ - \frac{\gamma}{2}\), so \(\tan\left(\frac{\alpha+\beta}{2}\right) = \cot\frac{\gamma}{2}\):

\frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}} = \frac{1}{\tan\frac{\gamma}{2}} \implies \tan\frac{\alpha}{2}\tan\frac{\gamma}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} = 1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}

\tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1

Answer: Both conditional identities are proved.

Example 10: Express \(3\sin\theta + 4\cos\theta\) in the form \(r\sin(\theta+\phi)\), where the terminal side of \(\phi\) is in Quadrant I

\(r = \sqrt{3^2+4^2} = 5\) nikalenge, phir \(\cos\phi = \frac{3}{5}\), \(\sin\phi = \frac{4}{5}\) se \(\phi\) maloom karenge.

r = \sqrt{9+16} = 5, \qquad \cos\phi = \frac{3}{5}, \quad \sin\phi = \frac{4}{5} \implies \phi = \tan^{-1}\frac{4}{3}

3\sin\theta + 4\cos\theta = 5\sin(\theta+\phi), \quad \phi = \tan^{-1}\frac{4}{3}

Answer: \(5\sin(\theta+\phi)\) with \(\phi = \tan^{-1}\frac{4}{3}\).

Example 11: Prove that \(\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta\)

Numerator aur denominator dono mein common factors nikalenge — \(\sin\theta(1+2\cos\theta)\) aur \(\cos\theta(1+2\cos\theta)\) banenge.

\text{L.H.S} = \frac{\sin\theta + 2\sin\theta\cos\theta}{(1+\cos 2\theta) + \cos\theta} = \frac{\sin\theta(1+2\cos\theta)}{2\cos^2\theta + \cos\theta} = \frac{\sin\theta(1+2\cos\theta)}{\cos\theta(1+2\cos\theta)} = \tan\theta = \text{R.H.S}

Answer: The identity is proved.

Example 12: Show that (i) \(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}\) and (ii) \(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\)

R.H.S mein tan ko sin/cos mein likh kar simplify karenge.

(i)

\text{R.H.S} = \frac{2\cdot\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}} = \frac{2\sin\theta\cos\theta}{\cos^2\theta+\sin^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta = \text{L.H.S}

(ii)

\text{R.H.S} = \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}} = \frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \cos^2\theta-\sin^2\theta = \cos 2\theta = \text{L.H.S}

Answer: Both formulas are proved.

Example 13: Reduce \(\cos^4\theta\) to an expression involving only functions of multiples of \(\theta\), raised to the first power

Power reducing formula \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) ko do martaba apply karenge.

\cos^4\theta = (\cos^2\theta)^2 = \left(\frac{1+\cos 2\theta}{2}\right)^2 = \frac{1}{4}\big(1 + 2\cos 2\theta + \cos^2 2\theta\big)

Using \(\cos^2 2\theta = \frac{1+\cos 4\theta}{2}\):

= \frac{1}{4}\left[1 + 2\cos 2\theta + \frac{1+\cos 4\theta}{2}\right] = \frac{3 + 4\cos 2\theta + \cos 4\theta}{8}

Answer: \(\cos^4\theta = \frac{3 + 4\cos 2\theta + \cos 4\theta}{8}\).

Example 14: Express \(2\sin 7\theta\cos 3\theta\) as a sum or difference

\(2\sin A\cos B = \sin(A+B) + \sin(A-B)\) ka formula lagayenge.

2\sin 7\theta\cos 3\theta = \sin(7\theta+3\theta) + \sin(7\theta-3\theta) = \sin 10\theta + \sin 4\theta

Answer: \(\sin 10\theta + \sin 4\theta\).

Example 15: Prove without using tables or a calculator that \(\sin 19^\circ\cos 11^\circ + \sin 71^\circ\sin 11^\circ = \frac{1}{2}\)

\(\sin 71^\circ = \sin(90^\circ-19^\circ) = \cos 19^\circ\) likhne par \(\sin(A+B)\) ka pattern ban jayega.

\text{L.H.S} = \sin 19^\circ\cos 11^\circ + \cos 19^\circ\sin 11^\circ = \sin(19^\circ+11^\circ) = \sin 30^\circ = \frac{1}{2} = \text{R.H.S}

Answer: The identity is proved.

Example 16: Express \(\sin 5x + \sin 7x\) as a product

\(\sin P + \sin Q = 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}\) ka formula lagayenge.

\sin 5x + \sin 7x = 2\sin\left(\frac{5x+7x}{2}\right)\cos\left(\frac{5x-7x}{2}\right) = 2\sin 6x\cos(-x) = 2\sin 6x\cos x

Answer: \(2\sin 6x\cos x\).

Example 17: Express \(\cos\theta + \cos 3\theta + \cos 5\theta + \cos 7\theta\) as a product

Terms ko pair kar ke sum-to-product formula lagayenge — pehle \((\cos\theta+\cos 7\theta)\) aur \((\cos 3\theta+\cos 5\theta)\), phir common factor nikalenge.

= (\cos\theta+\cos 7\theta) + (\cos 3\theta+\cos 5\theta) = 2\cos 4\theta\cos 3\theta + 2\cos 4\theta\cos\theta

= 2\cos 4\theta(\cos 3\theta + \cos\theta) = 2\cos 4\theta\cdot 2\cos 2\theta\cos\theta = 4\cos 4\theta\cos 2\theta\cos\theta

Answer: \(4\cos\theta\cos 2\theta\cos 4\theta\).

Example 18: Show that \(\cos 20^\circ\cos 40^\circ\cos 80^\circ = \frac{1}{8}\)

\(2\sin 20^\circ\) se multiply aur divide karenge — numerator mein lagatar \(\sin 2A\) banta jayega.

\text{L.H.S} = \frac{2\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 40^\circ\cos 40^\circ\cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 80^\circ\cos 80^\circ}{4\sin 20^\circ}

= \frac{\sin 160^\circ}{8\sin 20^\circ} = \frac{\sin(180^\circ-20^\circ)}{8\sin 20^\circ} = \frac{1}{8} = \text{R.H.S}

Answer: The identity is proved.

Mathematics 11 (Punjab Board)

MCQs: Chapter 10 – Trigonometric Identities

\(\cos(\alpha - \beta)\) equals:
1. \(\cos\alpha\cos\beta + \sin\alpha\sin\beta\)
2. \(\cos\alpha\cos\beta - \sin\alpha\sin\beta\)
3. \(\sin\alpha\cos\beta + \cos\alpha\sin\beta\)
4. \(\sin\alpha\cos\beta - \cos\alpha\sin\beta\)
Yeh Fundamental Law of Trigonometry hai — tamam compound angle identities isi se nikalti hain, is liye ise sab se pehle yaad karen.
\(\sin\left(\frac{\pi}{2} + \theta\right)\) equals:
1. \(\sin\theta\)
2. \(-\sin\theta\)
3. \(\cos\theta\)
4. \(-\cos\theta\)
\(\frac{\pi}{2}\) odd multiple hai is liye sin, cos mein badal jata hai, aur doosre quadrant mein sin positive hota hai — jawab \(\cos\theta\).
Which of the following is NOT equal to \(\cos 2\theta\)?
1. \(2\cos^2\theta - 1\)
2. \(1 - 2\sin^2\theta\)
3. \(\cos^2\theta - \sin^2\theta\)
4. \(2\sin\theta\cos\theta\)
Pehle teen options \(\cos 2\theta\) ke equivalent forms hain, jabke \(2\sin\theta\cos\theta = \sin 2\theta\) hota hai — is liye ye \(\cos 2\theta\) ke barabar nahi.
\(\tan(\alpha + \beta)\) equals:
1. \(\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)
2. \(\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}\)
3. \(\frac{\tan\alpha + \tan\beta}{1 + \tan\alpha\tan\beta}\)
4. \(\frac{\tan\alpha - \tan\beta}{1 - \tan\alpha\tan\beta}\)
Sum formula mein numerator mein plus aur denominator mein minus aata hai — ye \(\sin(\alpha+\beta)\) ko \(\cos(\alpha+\beta)\) se divide kar ke banta hai.
\(2\cos A\cos B\) equals:
1. \(\sin(A+B) + \sin(A-B)\)
2. \(\cos(A+B) + \cos(A-B)\)
3. \(\cos(A-B) - \cos(A+B)\)
4. \(\sin(A-B) - \sin(A+B)\)
\(\cos(A+B)\) aur \(\cos(A-B)\) ke formulas ko add karne par sin wale terms cancel ho jate hain aur \(2\cos A\cos B\) bach jata hai.
\(\cos(180^\circ - \theta)\) equals:
1. \(\cos\theta\)
2. \(-\cos\theta\)
3. \(\sin\theta\)
4. \(-\sin\theta\)
\(180^\circ\), \(\frac{\pi}{2}\) ka even multiple hai is liye ratio change nahi hoga, aur doosre quadrant mein cos negative hota hai.
\(\sin 3\theta\) equals:
1. \(3\sin\theta - 4\sin^3\theta\)
2. \(4\sin^3\theta - 3\sin\theta\)
3. \(3\sin\theta + 4\sin^3\theta\)
4. \(4\cos^3\theta - 3\cos\theta\)
Triple angle formula \(\sin(2\theta + \theta)\) ko expand kar ke banta hai — pehla term \(3\sin\theta\) aur doosra \(-4\sin^3\theta\) hota hai.
\(\tan(\alpha - \beta)\) equals:
1. \(\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)
2. \(\frac{\tan\alpha + \tan\beta}{1 + \tan\alpha\tan\beta}\)
3. \(\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}\)
4. \(\frac{\tan\alpha - \tan\beta}{1 - \tan\alpha\tan\beta}\)
Difference formula mein numerator mein minus aur denominator mein plus aata hai — sum formula ka ulta pattern hai.
\(\sin(\alpha - \beta)\) equals:
1. \(\sin\alpha\cos\beta - \cos\alpha\sin\beta\)
2. \(\sin\alpha\cos\beta + \cos\alpha\sin\beta\)
3. \(\cos\alpha\cos\beta - \sin\alpha\sin\beta\)
4. \(\cos\alpha\cos\beta + \sin\alpha\sin\beta\)
Sine ke difference formula mein darmiyan mein minus sign aata hai: \(\sin\alpha\cos\beta - \cos\alpha\sin\beta\).
The exact value of \(\sin 15^\circ\) is:
1. \(\frac{\sqrt{3}+1}{2\sqrt{2}}\)
2. \(\frac{1}{2}\)
3. \(\frac{\sqrt{3}}{2}\)
4. \(\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(15^\circ = 45^\circ - 30^\circ\) likh kar \(\sin(A-B)\) ka formula lagayen: \(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}\).
\(\cos 3\theta\) equals:
1. \(3\cos^3\theta - 4\cos\theta\)
2. \(4\cos^3\theta - 3\cos\theta\)
3. \(3\cos\theta - 4\cos^3\theta\)
4. \(4\sin^3\theta - 3\sin\theta\)
Cosine ka triple angle formula \(4\cos^3\theta - 3\cos\theta\) hai — sine wale formula ka ulta pattern (pehle 4, phir 3).
\(\sin\left(\frac{\pi}{2} - \theta\right)\) equals:
1. \(\cos\theta\)
2. \(-\cos\theta\)
3. \(\sin\theta\)
4. \(-\sin\theta\)
Odd multiple hone ki wajah se sin, cos mein badal jata hai aur pehle quadrant mein sign positive rehta hai.
\(\cos(A+B) + \cos(A-B)\) equals:
1. \(2\sin A\cos B\)
2. \(2\sin A\sin B\)
3. \(2\cos A\cos B\)
4. \(-2\sin A\sin B\)
Dono cosine formulas ko add karne par \(-\sin A\sin B\) aur \(+\sin A\sin B\) cancel ho jate hain.
\(\tan\left(\frac{\pi}{4} + \theta\right)\) equals:
1. \(\frac{1 + \tan\theta}{1 - \tan\theta}\)
2. \(\frac{1 - \tan\theta}{1 + \tan\theta}\)
3. \(\frac{\tan\theta - 1}{\tan\theta + 1}\)
4. \(1\)
\(\tan(A+B)\) mein \(A = \frac{\pi}{4}\) rakhen aur \(\tan\frac{\pi}{4} = 1\) use karen.
If \(\alpha + \beta + \gamma = 180^\circ\), then \(\tan\alpha + \tan\beta + \tan\gamma\) equals:
1. \(0\)
2. \(1\)
3. \(\tan\alpha + \tan\beta\)
4. \(\tan\alpha\tan\beta\tan\gamma\)
Triangle ke angles ke liye ye mashhoor conditional identity hai jo \(\tan(\alpha+\beta) = -\tan\gamma\) se sabit hoti hai.
\(\sin P + \sin Q\) equals:
1. \(2\cos\frac{P+Q}{2}\sin\frac{P-Q}{2}\)
2. \(2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}\)
3. \(2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}\)
4. \(-2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2}\)
Sum-to-product formula: sines ka sum, sum ke half-angle ka sine aur difference ke half-angle ka cosine hota hai.
The half-angle formula for \(\sin\frac{\alpha}{2}\) is:
1. \(\pm\sqrt{\frac{1-\cos\alpha}{2}}\)
2. \(\pm\sqrt{\frac{1+\cos\alpha}{2}}\)
3. \(\pm\sqrt{\frac{1-\sin\alpha}{2}}\)
4. \(\frac{1-\cos\alpha}{2}\)
\(\cos\alpha = 1 - 2\sin^2\frac{\alpha}{2}\) se derive hota hai; sign \(\frac{\alpha}{2}\) ke quadrant se decide hota hai.
\(\cos(270^\circ - \theta)\) equals:
1. \(\sin\theta\)
2. \(-\sin\theta\)
3. \(\cos\theta\)
4. \(-\cos\theta\)
\(270^\circ\), \(90^\circ\) ka odd multiple hai is liye cos, sin mein badlega, aur teesre quadrant mein cos negative hota hai.
\(\tan 2\theta\) equals:
1. \(\frac{2\tan\theta}{1+\tan^2\theta}\)
2. \(\frac{1-\tan^2\theta}{2\tan\theta}\)
3. \(\frac{2\tan\theta}{1-\tan^2\theta}\)
4. \(\tan^2\theta - 1\)
\(\tan(\theta+\theta)\) ka formula lagayen — numerator mein \(2\tan\theta\) aur denominator mein \(1-\tan^2\theta\).
To write \(a\sin\theta + b\cos\theta\) as \(r\sin(\theta+\phi)\), the value of \(r\) is:
1. \(\sqrt{a^2+b^2}\)
2. \(a+b\)
3. \(\sqrt{a^2-b^2}\)
4. \(|a-b|\)
\(a = r\cos\phi\) aur \(b = r\sin\phi\) ko square kar ke add karne par \(r^2 = a^2+b^2\) milta hai.
\(\cos^2\theta\) equals:
1. \(\frac{1-\cos 2\theta}{2}\)
2. \(\frac{1+\sin 2\theta}{2}\)
3. \(\frac{1-\sin 2\theta}{2}\)
4. \(\frac{1+\cos 2\theta}{2}\)
\(\cos 2\theta = 2\cos^2\theta - 1\) ko rearrange karen — power reduction ka ye formula \(\cos^4\theta\) jaisi powers reduce karne mein kaam aata hai.
The exact value of \(\sin 75^\circ\) is:
1. \(\frac{\sqrt{3}-1}{2\sqrt{2}}\)
2. \(\frac{\sqrt{3}+1}{2\sqrt{2}}\)
3. \(\frac{\sqrt{3}}{2}\)
4. \(1\)
\(75^\circ = 45^\circ + 30^\circ\) likh kar \(\sin(A+B)\) ka formula lagayen.

Mathematics 11 (Punjab Board)

Chapter 10: Trigonometric Identities — Formula Sheet

Fundamental Law of Trigonometry

\[ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \]

Sum and Difference Identities

\[ \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \]

\[ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \]

\[ \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \]

\[ \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} \]

\[ \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} \]

Allied Angles

\[ \sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta, \quad \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta, \quad \tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta \]

\[ \sin\left(\frac{\pi}{2} + \theta\right) = \cos\theta, \quad \cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta, \quad \tan\left(\frac{\pi}{2} + \theta\right) = -\cot\theta \]

\[ \sin(\pi - \theta) = \sin\theta, \quad \cos(\pi - \theta) = -\cos\theta, \quad \tan(\pi - \theta) = -\tan\theta \]

\[ \sin(\pi + \theta) = -\sin\theta, \quad \cos(\pi + \theta) = -\cos\theta, \quad \tan(\pi + \theta) = \tan\theta \]

\[ \sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta, \quad \cos\left(\frac{3\pi}{2} - \theta\right) = -\sin\theta, \quad \tan\left(\frac{3\pi}{2} - \theta\right) = \cot\theta \]

\[ \sin\left(\frac{3\pi}{2} + \theta\right) = -\cos\theta, \quad \cos\left(\frac{3\pi}{2} + \theta\right) = \sin\theta, \quad \tan\left(\frac{3\pi}{2} + \theta\right) = -\cot\theta \]

\[ \sin(2\pi - \theta) = -\sin\theta, \quad \cos(2\pi - \theta) = \cos\theta, \quad \tan(2\pi - \theta) = -\tan\theta \]

Double Angle Identities

\[ \sin 2\alpha = 2\sin\alpha\cos\alpha = \frac{2\tan\alpha}{1+\tan^2\alpha} \]

\[ \cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha = \frac{1-\tan^2\alpha}{1+\tan^2\alpha} \]

\[ \tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2\alpha} \]

Half Angle Identities

\[ \sin\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{2}}, \quad \cos\frac{\alpha}{2} = \pm\sqrt{\frac{1+\cos\alpha}{2}}, \quad \tan\frac{\alpha}{2} = \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}} \]

Triple Angle Identities

\[ \sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha \]

\[ \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha \]

\[ \tan 3\alpha = \frac{3\tan\alpha - \tan^3\alpha}{1-3\tan^2\alpha} \]

Power Reduction

\[ \sin^2\alpha = \frac{1-\cos 2\alpha}{2}, \quad \cos^2\alpha = \frac{1+\cos 2\alpha}{2} \]

\[ \sin^4\theta = \frac{3 - 4\cos 2\theta + \cos 4\theta}{8}, \quad \cos^4\theta = \frac{3 + 4\cos 2\theta + \cos 4\theta}{8} \]

Product to Sum

\[ 2\sin\alpha\cos\beta = \sin(\alpha+\beta) + \sin(\alpha-\beta) \]

\[ 2\cos\alpha\sin\beta = \sin(\alpha+\beta) - \sin(\alpha-\beta) \]

\[ 2\cos\alpha\cos\beta = \cos(\alpha+\beta) + \cos(\alpha-\beta) \]

\[ -2\sin\alpha\sin\beta = \cos(\alpha+\beta) - \cos(\alpha-\beta) \]

Sum to Product

\[ \sin P + \sin Q = 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2} \]

\[ \sin P - \sin Q = 2\cos\frac{P+Q}{2}\sin\frac{P-Q}{2} \]

\[ \cos P + \cos Q = 2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2} \]

\[ \cos P - \cos Q = -2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2} \]

Auxiliary Angle

\[ a\sin\theta + b\cos\theta = r\sin(\theta + \phi), \quad r = \sqrt{a^2+b^2}, \quad \tan\phi = \frac{b}{a} \]

Useful Factorizations

\[ \sin(\alpha+\beta)\sin(\alpha-\beta) = \sin^2\alpha - \sin^2\beta = \cos^2\beta - \cos^2\alpha \]

\[ \cos(\alpha+\beta)\cos(\alpha-\beta) = \cos^2\alpha - \sin^2\beta = \cos^2\beta - \sin^2\alpha \]

\[ 1 + \sin 2\theta = (\sin\theta + \cos\theta)^2, \quad 1 - \sin 2\theta = (\sin\theta - \cos\theta)^2 \]

\[ 1 + \cos 2\alpha = 2\cos^2\alpha, \quad 1 - \cos 2\alpha = 2\sin^2\alpha \]

Conditional Identities (Triangle: \(\alpha+\beta+\gamma = 180^\circ\))

\[ \tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma \]

\[ \tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1 \]

\[ \cot\frac{\alpha}{2} + \cot\frac{\beta}{2} + \cot\frac{\gamma}{2} = \cot\frac{\alpha}{2}\cot\frac{\beta}{2}\cot\frac{\gamma}{2} \]

Chapter 10 FAQs

What is covered in Chapter 10 (Trigonometric Identities)?

This chapter covers Sum and difference identities, double and half angle formulas, sum-to-product. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 10 carry in the Punjab Board paper?

Chapter 10 (Trigonometric Identities) carries approximately 12 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 10?

Use the exercise index on this page — every exercise of Chapter 10 has its own page with complete step-by-step solutions.