12. Multiple and Submultiple Angles#

12.1. Multiple Angles#

An angle of the form nA,nA, where nn is an integer is called a multiple angle. For example, 2A,3A,4A,…2A, 3A, 4A, \ldots are multiple angles of A.A.

12.1.1. Trigonometrical Ratios of 2A2A#

From previous chapter we know that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A\cos B + \cos A\sin B

Substituting B=A,B = A, we get

sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A

Similarly, cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos 2A = \cos^2A - \sin^2A = 2\cos^2A - 1 = 1 - 2\sin^2A (recall formula from previous chapter and substitute B=AB = A cos⁡2A=1−sin⁡2A\cos^2A = 1 -\sin^2A and sin⁡2A=1−cos⁡2a\sin^2A = 1 - \cos^2a)

Also, tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \frac{2\tan A}{1 - \tan^2A} (recall formula from previous chapter and put B=AB = A)

12.1.2. sin⁡2A\sin 2A and cos⁡2A\cos 2A in terms of tan⁡A\tan A#

sin⁡2A=2sin⁡Acos⁡Asin⁡2A+cos⁡2A[∵sin⁡2A+cos⁡2A=1]\sin 2A = \frac{2\sin A\cos A}{\sin^2A + \cos^2A}[\because \sin^2A + \cos^2A = 1]

Dividing both numerator and denominator by cos⁡2A,\cos^2A, we get

sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A = \frac{2\tan A}{1 + \tan^2A}

cos⁡A=cos⁡2A−sin⁡2A=cos⁡2A−sin⁡2Acos⁡2A+sin⁡2A[∵sin⁡2A+cos⁡2A=1]\cos A = \cos^2A - \sin^2A = \frac{\cos^2A - \sin^2A}{\cos^2A + \sin^2A}[\because \sin^2A + \cos^2A = 1]

Dividing both numerator and denominator by cos⁡2A,\cos^2A, we get

cos⁡2A=1−tan⁡2A1+tan⁡2A=cot⁡2A−1cot⁡2A+1\cos 2A = \frac{1 - \tan^2A}{1 + \tan^2A} = \frac{\cot^2A - 1}{\cot^2A + 1}

12.1.3. Trigonometrical Ratios of 3A3A#

sin⁡3A=sin⁡2Acos⁡A+cos⁡2Asin⁡A=2sin⁡Acos⁡2A+cos⁡2Asin⁡A−sin⁡3A\sin 3A = \sin2A\cos A + \cos 2A\sin A = 2\sin A\cos^2 A + \cos^2A\sin A - \sin^3A

=2sin⁡A(1−sin⁡2A)+(1−2sin⁡2A)sin⁡A−sin⁡3A= 2\sin A(1 - \sin^2A) + (1 - 2\sin^2A)\sin A - \sin^3A

=3sin⁡A−4sin⁡3A= 3\sin A - 4\sin^3A

cos⁡3A=cos⁡2Acos⁡A−sin⁡2Asin⁡A=(2cos⁡2A−1)cos⁡A−2sin⁡2Acos⁡A\cos 3A = \cos2A\cos A - \sin 2A\sin A = (2\cos^2A - 1)\cos A - 2\sin^2 A\cos A

=2cos⁡3A−cos⁡A−2(1−cos⁡2A)cos⁡A= 2\cos^3A - \cos A - 2(1 - \cos^2A)\cos A

=4cos⁡3A−3cos⁡A= 4\cos^3 A - 3\cos A

We know that tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}

Putting B=AB = A and C=A,C = A, we get

tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A = \frac{3\tan A - \tan^3A}{1 - 3\tan^2A}

Similarly cot⁡3A=cot⁡3A−3cot⁡A3cot⁡2A−1\cot 3A = \frac{\cot^3 A - 3\cot A}{3\cot^2A - 1}

12.1.4. Some Important Formulae#

  1. cos⁡2A=1−2sin⁡2A⇒sin⁡2A=12(1−cos⁡2A)\cos2A = 1 - 2\sin^2A \Rightarrow \sin^2A = \frac{1}{2}(1 - \cos2A)

  2. cos⁡2A=2cos⁡2−1⇒cos⁡2A=12(1+cos⁡2A)\cos2A = 2\cos^2 - 1 \Rightarrow \cos^2A = \frac{1}{2}(1 + \cos2A)

  3. sin⁡3A=3sin⁡A−4sin⁡3A⇒sin3A=12(3sin⁡A−sin⁡3A)\sin 3A = 3\sin A - 4\sin^3A \Rightarrow sin^3A = \frac{1}{2}(3\sin A - \sin3A)

  4. cos⁡3A=4cos⁡3A−3cos⁡A⇒cos⁡3A=14(cos⁡3A+3cos⁡A)\cos 3A = 4\cos^3A - 3\cos A \Rightarrow \cos^3A = \frac{1}{4}(\cos3A + 3\cos A)

12.2. Submultiple Angles#

An angle of the form An,\frac{A}{n}, where nn is an integer is called a submultiple angle. For exmaple, A2,A3,A4,…\frac{A}{2}, \frac{A}{3}, \frac{A}{4}, \ldots are submultiple angles of A.A.

12.2.1. Trigonometrical Ratios of A/2A/2#

We know that, sin⁡2A=2sin⁡Acos⁡A.\sin 2A = 2\sin A\cos A. Putting A=A/2,A=A/2, we get

sin⁡A=2sin⁡A/2cos⁡A/2\sin A = 2\sin A/2\cos A/2

cos⁡2A=cos⁡2A−sin⁡A.\cos 2A = \cos^2A - \sin^A. Putting A=A/2,A = A/2, we get

cos⁡A−cos⁡2A2−sin⁡2A2\cos A - \cos^2\frac{A}{2} - \sin^2\frac{A}{2}

cos⁡2A=2cos⁡2A−1.\cos 2A = 2\cos^2A - 1. Putting A=A/2,A = A/2, we get

cos⁡A=2cos⁡2A2−1\cos A = 2\cos^2\frac{A}{2} - 1

cos⁡2A=1−2sin⁡2A.\cos 2A = 1 - 2\sin^2A. Putting A=A/2,A = A/2, we get

cos⁡A=1−2sin⁡2A2\cos A = 1 - 2\sin^2\frac{A}{2}

tan⁡2A=2tan⁡A1−tan⁡2A.\tan 2A = \frac{2\tan A}{1 - \tan^2A}. Putting A=A/2,A = A/2, we get

tan⁡A=2tan⁡A21−tan⁡2A2\tan A = \frac{2\tan \frac{A}{2}}{1 - \tan^2\frac{A}{2}}

sin⁡2A=2tan⁡A1+tan⁡2A∴sin⁡A=2tan⁡A21+tan⁡2A2\sin 2A = \frac{2\tan A}{1 + \tan^2A} \therefore \sin A = \frac{2\tan \frac{A}{2}}{1 + \tan^2\frac{A}{2}}

cos⁡2A=1−tan⁡2A1+tan⁡2A∴cos⁡A=1−tan⁡2A21+tan⁡2A2\cos 2A = \frac{1 - \tan^2A}{1 + \tan^2A} \therefore \cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}

cot⁡2A=cot⁡2A−12cot⁡A∴cot⁡A=cot⁡2A2−12cot⁡A2\cot 2A = \frac{\cot^2A - 1}{2\cot A} \therefore \cot A = \frac{\cot^2\frac{A}{2} - 1}{2\cot \frac{A}{2}}

12.2.2. Trigonometrical Ratios of A/3A/3#

sin⁡3A=3sin⁡A−4sin⁡3A.\sin 3A = 3\sin A - 4\sin^3A. Putting A=A3,A = \frac{A}{3}, we get

sin⁡A=3sin⁡A3−4sin⁡3A3\sin A = \frac{3}{\sin \frac{A}{3}} - 4\sin^3\frac{A}{3}

cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4\cos^3A - 3\cos A. Putting A=A3,A = \frac{A}{3}, we get

cos⁡A=4cos⁡3A3−3cos⁡A3\cos A = 4\cos^3\frac{A}{3} - 3\cos \frac{A}{3}

tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A = \frac{3\tan A - \tan^3A}{1 - 3\tan^2A}

tan⁡A=3tan⁡A3−tan⁡3A31−3tan⁡2A3\tan A = \frac{3\tan\frac{A}{3} - \tan^3\frac{A}{3}}{1 - 3\tan^2\frac{A}{3}}

12.2.3. Values of cosA/2,sin⁡A/2cos A/2, \sin A/2 and tan⁡A/2\tan A/2 in terms of cos⁡A\cos A#

cos⁡2A2=1+cos⁡A2∴cos⁡A2=1+cos⁡A2\cos^2\frac{A}{2} = \frac{1 + \cos A}{2} \therefore \cos \frac{A}{2} = \sqrt{\frac{1 + \cos A}{2}}

sin⁡2A2=1−cos⁡A2∴sin⁡A2=1−cos⁡A2\sin^2\frac{A}{2} = \frac{1 - \cos A}{}2 \therefore \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}

tan⁡2A2=1−cos⁡A1+cos⁡A∴tan⁡A2=1−cos⁡A1+cos⁡A\tan^2\frac{A}{2} = \frac{1 - \cos A}{1 + \cos A} \therefore \tan\frac{A}{2} = \sqrt{\frac{1 - \cos A}{1 + \cos A}}

12.2.4. Values of sin⁡A/2\sin A/2 and cos⁡A/2\cos A/2 in terms of sin⁡A\sin A#

(cos⁡A2+sin⁡A2)2=cos⁡2A2+sin⁡2A2+2cos⁡A2sin⁡A2\left(\cos \frac{A}{2} + \sin\frac{A}{2}\right)^2 = \cos^2\frac{A}{2} + \sin^2\frac{A}{2} + 2\cos\frac{A}{2}\sin\frac{A}{2}

=1+sin⁡A⇒cos⁡A2+sin⁡A2=1+sin⁡A= 1 + \sin A \Rightarrow \cos \frac{A}{2} + \sin \frac{A}{2} = \sqrt{1 + \sin A}

Similarly, cos⁡A2−sin⁡A2=1−sin⁡A\cos \frac{A}{2} - \sin \frac{A}{2} = \sqrt{1 - \sin A}

Adding, we get cos⁡A2=±121+sin⁡A±121−sin⁡A\cos \frac{A}{2} = \pm\frac{1}{2}\sqrt{1 + \sin A} \pm\frac{1}{2}\sqrt{1 - \sin A}

Subtracting, we get cos⁡A2=±121+sin⁡A∓121−sin⁡A\cos \frac{A}{2} = \pm\frac{1}{2}\sqrt{1 + \sin A} \mp\frac{1}{2}\sqrt{1 - \sin A}

12.2.5. Value of sin⁡18∘\sin 18^\circ and cos⁡72∘\cos 72^\circ#

Let A=18∘,A = 18^\circ, then sin⁡5A=90∘∴2A+3A=90∘\sin 5A = 90^\circ \therefore 2A + 3A = 90^\circ

sin⁡2A=sin⁡(90∘−sin⁡3A)∴2sin⁡Acos⁡A=4cos⁡3A−3cos⁡A\sin2A = \sin(90^\circ - \sin 3A) \therefore 2\sin A\cos A = 4\cos^3A - 3\cos A

Dividing both sides by cos⁡A,\cos A, we get

2sin⁡A=4cos⁡2A−3=4(1−sin⁡2A)−32\sin A = 4\cos^2A - 3 = 4(1 - \sin^2A) - 3

4sin⁡2A+2sin⁡A−1=04\sin^2A + 2\sin A - 1 = 0

sin⁡A=−1±54\sin A = \frac{-1\pm\sqrt{5}}{4}

However, since A=18∘∴sin⁡A>0A= 18^\circ\therefore \sin A > 0

∴sin⁡18∘=−1+54\therefore \sin18^\circ = \frac{-1 + \sqrt{5}}{4}

∴sin⁡(90∘−18∘)=cos⁡72∘=5−14\therefore \sin(90^\circ - 18^\circ) = \cos72^\circ = \frac{\sqrt{5} - 1}{4}

12.2.6. Value of cos⁡18∘\cos 18^\circ and sin⁡72∘\sin 72^\circ#

cos⁡218∘=1−sin⁡218∘=1−(5−14)2\cos^218^\circ = 1 - \sin^218^\circ = 1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2

=10+2516∴cos⁡18∘=1410+25[∵cos⁡18∘>0]= \frac{10 + 2\sqrt{5}}{16}\therefore \cos18^\circ = \frac{1}{4}\sqrt{10 + 2\sqrt{5}}[\because \cos18^\circ > 0]

cos⁡(90∘−18∘)=sin⁡72∘=1410+25\cos(90^\circ - 18^\circ) = \sin72^\circ = \frac{1}{4}\sqrt{10 + 2\sqrt{5}}

12.2.7. Value of tan⁡18∘\tan 18^\circ and tan⁡72∘\tan 72^\circ#

tan⁡18∘=sin⁡18∘cos⁡18∘=5−110+25\tan 18^\circ = \frac{\sin18^\circ}{\cos18^\circ} = \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}

tan⁡18∘cot⁡18∘=1⇒tan⁡72∘=1tan⁡18∘=10+255−1\tan18^\circ\cot18^\circ = 1\Rightarrow \tan72^\circ = \frac{1}{\tan18^\circ} = \frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{5} - 1} x Value of cos⁡36∘\cos 36^\circ and sin⁡54∘\sin 54^\circ ——————————————————– cos⁡36∘=1−2sin⁡218∘=1−2(5−14)2\cos 36^\circ = 1 - 2\sin^218^\circ = 1 - 2\left(\frac{\sqrt{5} - 1}{4}\right)^2

=5+14= \frac{\sqrt{5} + 1}{4}

sin⁡54∘=sin⁡(90∘−36∘)=cos⁡36∘=5+14\sin 54^\circ = \sin(90^\circ - 36^\circ) = \cos36^\circ = \frac{\sqrt{5} + 1}{4}

12.2.8. Value of sin⁡36∘\sin 36^\circ and cos⁡54∘\cos 54^\circ#

sin⁡36∘=1−cos⁡236∘=1−(5+14)2\sin36^\circ = 1 - \cos^236^\circ = 1 - \left(\frac{\sqrt{5} + 1}{4}\right)^2

=1410−25= \frac{1}{4}\sqrt{10 - 2\sqrt{5}}

cos⁡54∘=cos⁡(90∘−36∘)=sin⁡36∘=1410−25\cos54^\circ = \cos(90^\circ - 36^\circ) = \sin36^\circ = \frac{1}{4}\sqrt{10 - 2\sqrt{5}}

Several other angles like, 9∘,15∘,2212∘,712∘9^\circ, 15^\circ, 22\frac{1}{2}^\circ, 7\frac{1}{2}^\circ etc can be found similarrly.

12.3. Problems#

  1. Find the value of sin⁡2A,\sin 2A, when

    1. cos⁡A=35\cos A = \frac{3}{5}

    2. sin⁡A=1213\sin A = \frac{12}{13}

    3. tan⁡A=1663\tan A = \frac{16}{63}

  2. Find the value of cos⁡2A,\cos 2A, when

    1. cos⁡A=1517\cos A = \frac{15}{17}

    2. sin⁡A=45\sin A = \frac{4}{5}

    3. tan⁡A=512\tan A = \frac{5}{12}

  3. If tan⁡A=ba,\tan A = \frac{b}{a}, find the value of acos⁡2A+bsin⁡2Aa\cos 2A+ b\sin 2A

Prove that

  1. sin⁡2A1+cos⁡2A=tan⁡A\frac{\sin 2A}{1 + \cos 2A} = \tan A

  2. sin⁡2A1−cos⁡2A=cot⁡A\frac{\sin 2A}{1 - \cos 2A} = \cot A

  3. 1−cos⁡2A1+cos⁡2A=tan⁡2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A

  4. tan⁡A+cot⁡A=2cosec⁡2A\tan A + \cot A = 2\cosec 2A

  5. tan⁡A−cot⁡A=−2cot⁡2A\tan A - \cot A = -2\cot2A

  6. cosec⁡2A+cot⁡2A=cot⁡A\cosec 2A + \cot 2A = \cot A

  7. 1−cos⁡A+cos⁡B−cos⁡(A+B)1+cos⁡A−cos⁡B−cos⁡(A+B)=tan⁡A2cot⁡B2\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)} = \tan\frac{A}{2}\cot\frac{B}{2}

  8. cos⁡A1∓sin⁡A=tan⁡(45∘±A2)\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)

  9. sec⁡8A−1sec⁡4A−1=tan⁡8Atan⁡2A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}

  10. 1+tan⁡2(45∘−A)1−tan⁡2(45∘−A)=cosec⁡2A\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A

  11. sin⁡A+sin⁡Bsin⁡A−sin⁡B=tan⁡A+B2tan⁡A−B2\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}}

  12. sin⁡2A−sin⁡2Bsin⁡Acos⁡A−sin⁡Bcos⁡B=tan⁡(A+B)\frac{\sin^2A - \sin^2B}{\sin A\cos A - \sin B\cos B} = \tan(A + B)

  13. tan⁡(π4+A)−tan⁡(π4−A)=2tan⁡2A\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A

  14. cos⁡A+sin⁡Acos⁡A−sin⁡A−cos⁡A−sin⁡Acos⁡A+sin⁡A=2tan⁡2A\frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} = 2\tan 2A

  15. cot⁡(A+15∘)−tan⁡(A−15∘)=4cos⁡2A1+2sin⁡2A\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}

  16. sin⁡A+sin⁡2A1+cos⁡A+cos⁡2A=tan⁡A\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A

  17. 1+sin⁡A−cos⁡A1+sin⁡A+cosA=tan⁡A2\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}

  18. sin⁡(n+1)A−sin⁡(n−1)Acos⁡(n+1)A+2cos⁡nA+cos⁡(n−1)A=tan⁡A2\frac{\sin(n + 1)A - \sin(n - 1)A}{\cos(n + 1)A + 2\cos nA + \cos(n - 1)A} = \tan \frac{A}{2}

  19. sin⁡(n+1)A+2sin⁡nA+sin⁡(n−1)Acos⁡(n−1)−cos⁡(n+1)A=cot⁡A2\frac{\sin(n + 1)A + 2\sin nA + \sin(n - 1)A}{\cos(n - 1) - \cos(n + 1)A} = \cot \frac{A}{2}

  20. sin⁡(2n+1)Asin⁡A=sin⁡2(n+1)A−sin⁡2nA\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA

  21. sin⁡(A+3B)+sin⁡(3A+B)sin⁡2A+sin⁡2B=2cos⁡(A+B)\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)

  22. sin⁡3A+sin⁡2A−sin⁡A=4sin⁡Acos⁡A2cos⁡3A2\sin 3A + \sin 2A - \sin A = 4\sin A\cos \frac{A}{2}\cos \frac{3A}{2}

  23. tan⁡2A=(sec⁡2A+1)sec⁡2A−1\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}

  24. cos⁡32A+3cos⁡2A=4(cos⁡6A−sin⁡6A)\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)

  25. 1+cos⁡22A=2(cos⁡4A+sin⁡4A)1 + \cos^22A = 2(\cos^4A + \sin^4A)

  26. sec⁡2A(1+sec⁡2A)=2sec⁡2A\sec^2A(1 + \sec2A) = 2\sec2A

  27. cosec⁡A−2cot⁡2Acos⁡A=2sin⁡A\cosec A - 2\cot 2A\cos A = 2\sin A

  28. cot⁡A=12(cot⁡A2−tan⁡A2)\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)

  29. sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

  30. cos⁡Acos⁡(60∘−A)cos⁡(60∘+A)=14cos⁡3A\cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A

  31. cot⁡A+cot⁡(60∘+A)−cot⁡(60∘−A)=3cot⁡3A\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A

  32. cos⁡4A=1−8cos⁡2A+8cos⁡4A\cos 4A = 1 - 8\cos^2A + 8\cos^4A

  33. sin⁡4A=4sin⁡Acos⁡3A−4cos⁡Asin⁡3A\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A

  34. cos⁡6A=32cos⁡6A−48cos⁡4A+18cos⁡2A−1\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1

  35. tan⁡3Atan⁡2Atan⁡A=tan⁡3A−tan⁡2A−tan⁡A\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A

  36. 2cos⁡2nA+12cos⁡A+1=(2cos⁡A−1)(2cos⁡2A−1)(2cos⁡22A−1)…(2cos⁡2n−1−1)\frac{2\cos2^nA + 1}{2\cos A + 1} = (2\cos A - 1)(2\cos 2A - 1)(2\cos2^2A - 1)\ldots(2\cos2^{n - 1} - 1)

  37. If tan⁡A=17,sin⁡B=110,\tan A= \frac{1}{7}, \sin B = \frac{1}{\sqrt{10}}, prove that A+2B=π4,A + 2B = \frac{\pi}{4}, where 0<A<π40 < A < \frac{\pi}{4} and 0<B<π40 < B < \frac{\pi}{4}

Prove that

  1. tan⁡(π4+A)+tan⁡(π4−A)=2sec⁡2A\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A

  2. 3cosec⁡20∘−sec⁡20∘=4\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4

  3. tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

  4. cos⁡2A+cos⁡2(2π3−A)+cos⁡2(2π3+A)=32\cos^2A + \cos^2\left(\frac{2\pi}{3} - A\right) + \cos^2\left(\frac{2\pi}{3} + A\right) = \frac{3}{2}

  5. 2sin⁡2A+4cos⁡(A+B)sin⁡Asin⁡B+cos⁡2(A+B)2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B) is idnependent of A.A.

  6. If cos⁡A=12(a+1a),\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right), show that cos⁡2A=12(a2+1a2)\cos 2A = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

    Prove that

  7. cos⁡2A+sin⁡2Acos⁡2B=cos⁡2B+sin⁡2Bcos⁡2A\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A

  8. 1+tan⁡Atan⁡2A=sec⁡2A1 + \tan A\tan 2A = \sec 2A

  9. 1+sin⁡2A1−sin⁡2A=(1+tan⁡A1−tan⁡A)2\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2

  10. 1sin⁡10∘−3cos⁡10∘=4\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4

  11. cot⁡2A−tan⁡2A=4cot⁡2Acosec⁡2A\cot^2A - \tan^2A = 4\cot2A\cosec 2A

  12. 1+sin⁡2Acos⁡2A=cos⁡A+sin⁡Acos⁡A−sin⁡A=tan⁡(π4+A)\frac{1 +\sin 2A}{\cos2A} = \frac{\cos A + \sin A}{\cos A - \sin A} = \tan\left(\frac{\pi}{4} + A\right)

  13. cos⁡6A−sin⁡6A=cos⁡2A(1−14sin⁡22A)\cos^6A - \sin^6A = \cos2A\left(1 - \frac{1}{4}\sin^22A\right)

  14. cos⁡2A+cos⁡2(π3+A)+cos⁡2(π3−A)=32\cos^2A + \cos^2\left(\frac{\pi}{3} + A\right) + \cos^2\left(\frac{\pi}{3} - A\right)= \frac{3}{2}

  15. (1+sec⁡2A)(1+sec⁡22A)(1+sec⁡23A)…(1+sec⁡2nA)=tan⁡2nAtan⁡A(1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA) = \frac{\tan2^nA}{\tan A}

  16. sin⁡2nAsin⁡A=2ncos⁡Acos⁡2Acos⁡22A…cos⁡2n−1A\frac{\sin2^nA}{\sin A} = 2^n\cos A\cos 2A\cos 2^2A\ldots\cos2^{n - 1}A

  17. 3(sin⁡A−cos⁡A)4+6(sin⁡A+cos⁡A)2+4(sin⁡6A+cos⁡6A)=133(\sin A - \cos A)^4 + 6(\sin A + \cos A)^2 + 4(\sin^6A + \cos^6A) = 13

  18. 2(sin⁡6A+cos⁡6A)−3(sin⁡4A+cos⁡4A)+1=02(\sin^6A + \cos^6A) - 3(\sin^4A + \cos^4A) + 1 = 0

  19. cos⁡2A+cos⁡2(A+B)−2cos⁡Acos⁡Bcos⁡(A+B)\cos^2A + \cos^2(A + B) -2\cos A\cos B\cos(A + B) if independent of A.A.

  20. cos⁡3Acos⁡3A+sin⁡3Asin⁡3A=cos⁡32A\cos^3A\cos 3A + \sin^3A\sin 3A = \cos^32A

  21. tan⁡Atan⁡(60∘−A)tan⁡(60∘+A)=tan⁡3A\tan A\tan(60^\circ - A)\tan(60^\circ + A) = \tan 3A

  22. sin⁡2A+sin⁡3(2π3+A)+sin⁡3(4π3+A)=−34sin⁡3A\sin^2A + \sin^3\left(\frac{2\pi}{3} + A\right) + \sin^3\left(\frac{4\pi}{3} + A\right) = -\frac{3}{4}\sin 3A

  23. 4(cos⁡310∘+sin⁡320∘)=3(cos⁡10∘+sin⁡20∘)4(\cos^310^\circ + \sin^320^\circ) = 3(\cos 10\circ + \sin 20^\circ)

  24. sin⁡Acos⁡3A−cos⁡Asin⁡3A=14sin⁡4A\sin A\cos^3A - \cos A\sin^3A = \frac{1}{4}\sin 4A

  25. cos⁡3Asin⁡3A+sin⁡3Acos⁡3A=34sin⁡4A\cos^3A\sin3A + \sin^3A\cos 3A = \frac{3}{4}\sin 4A

  26. sin⁡Asin⁡(60∘+A)sin⁡(A+120∘)=sin⁡3A\sin A\sin(60^\circ + A)\sin(A + 120^\circ) = \sin 3A

  27. cot⁡A+cot⁡(60∘+A)+cot⁡(120∘+A)=3cot⁡3A\cot A + \cot(60^\circ + A) + \cot(120^\circ + A) = 3\cot 3A

  28. cos⁡5A=16cos⁡5A−20cos⁡3A+5cos⁡A\cos 5A = 16\cos^5A - 20\cos^3A + 5\cos A

  29. sin⁡5A=5sin⁡A−20sin⁡3A+16sin⁡5A\sin 5A = 5\sin A - 20\sin^3A + 16\sin^5A

  30. cos⁡4A−cos⁡4B=8(cos⁡A−cos⁡B)(cos⁡A+cos⁡B)(cos⁡A−sin⁡B)(cos⁡A+sin⁡B)\cos 4A - \cos 4B = 8(\cos A - \cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)

  31. tan⁡4A=4tan⁡A−4tan⁡3A1−6tan⁡2A+tan⁡4A\tan 4A = \frac{4\tan A - 4\tan^3A}{1 - 6\tan^2A + \tan^4A}

  32. If 2tan⁡A=3tan⁡B,2\tan A = 3\tan B, prove that tan⁡(A−B)=sin⁡2B5−cos⁡2B\tan (A- B) = \frac{\sin 2B}{5 - \cos 2B}

  33. If sin⁡A+sin⁡B=x\sin A + \sin B = x and cos⁡A+cos⁡B=y,\cos A + \cos B = y, show that sin⁡(A+B)=2xyx2+y2\sin(A + B) = \frac{2xy}{x^2 + y^2}

  34. If A=π2n+1,A= \frac{\pi}{2^n + 1}, prove that cos⁡A.cos⁡2A.cos⁡22A.….cos⁡2n−1A=12n\cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A = \frac{1}{2^n}

  35. If tan⁡A=yx,\tan A = \frac{y}{x}, prove that xcos⁡2A+ysin⁡2A=xx\cos 2A + y\sin 2A = x

  36. If tan⁡2A=1+2tan⁡2B,\tan^2A = 1 + 2\tan^2B, prove that cos⁡2B=1+2cos⁡2A\cos 2B = 1 + 2\cos 2A

  37. If AA and BB lie between 00 and π2\frac{\pi}{2} and cos⁡2A=3cos⁡2B−13−cos⁡2B,\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B}, prove that tan⁡A=2tan⁡B\tan A = \sqrt{2}\tan B

  38. If tan⁡B=3tan⁡A,\tan B = 3\tan A, prove that tan⁡(A+B)=2sin⁡2B1+cos⁡2B\tan(A + B) = \frac{2\sin 2B}{1 + \cos 2B}

  39. If xsin⁡A=ycos⁡A,x\sin A = y\cos A, prove that xsec⁡2A+ycosec⁡2A=x\frac{x}{\sec 2A} + \frac{y}{\cosec 2A} = x

  40. If tan⁡A=sec⁡2B,\tan A = \sec 2B, prove that sin⁡2A=1−tan⁡4B1+tan⁡4B\sin 2A = \frac{1 - \tan^4B}{1 + \tan^4B}

  41. If A=π3,A = \frac{\pi}{3}, prove that cos⁡A.cos⁡2A.cos⁡3A.cos⁡4A.cos⁡5A.cos⁡6A=−116\cos A.\cos 2A. \cos 3A.\cos 4A.\cos 5A.\cos 6A = -\frac{1}{16}

  42. If A=π15,A = \frac{\pi}{15}, prove that cos⁡2A.cos⁡4A.cos⁡8A.cos⁡14A=116\cos2A.\cos4A.\cos8A.\cos14A = \frac{1}{16}

  43. If tan⁡Atan⁡B=a−ba+b,\tan A\tan B = \sqrt{\frac{a - b}{a + b}}, prove that (a−bcos⁡2A)(a−bcos⁡2B)=a2−b2(a - b\cos2A)(a - b\cos2B) = a^2 - b^2

  44. If sin⁡A=12\sin A = \frac{1}{2} and sin⁡B=13,\sin B = \frac{1}{3}, find the value of sin⁡(A+B)\sin(A + B) and sin⁡(2A+2B)\sin(2A + 2B)

  45. If cos⁡A=1161\cos A = \frac{11}{61} and sin⁡B=45,\sin B = \frac{4}{5}, find the value of sin⁡2A−B2\sin^2 \frac{A - B}{2} and cos2A+B2,cos^2\frac{A + B}{2}, the angle of AA and BB being positive acute angles.

  46. Given sec⁡A=54,\sec A = \frac{5}{4}, find tan⁡A2\tan\frac{A}{2} and tan⁡A.\tan A.

  47. If cos⁡A=.3,\cos A = .3, find the value of tan⁡A2,\tan \frac{A}{2}, and explain the resulting ambiguity.

  48. If sin⁡A+sin⁡B=x\sin A + \sin B = x and cos⁡A+cos⁡B=y,\cos A + \cos B = y, find the value of tan⁡A−B2\tan \frac{A - B}{2}

Prove that

  1. (cos⁡A+cos⁡B)2+(sin⁡A−sin⁡B)2=4cos⁡2A+B2(\cos A + \cos B)^2 + (\sin A - \sin B)^2 = 4\cos^2 \frac{A + B}{2}

  2. (cos⁡A+cos⁡B)2+(sin⁡A+sin⁡B)2=4cos⁡2A−B2(\cos A + \cos B)^2 + (\sin A + \sin B)^2 = 4\cos^2 \frac{A - B}{2}

  3. (cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2=4sin⁡2A−B2(\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 4\sin^2 \frac{A - B}{2}

  4. sin⁡2(π8+A2)−sin⁡2(π8−A2)=12sin⁡A\sin^2\left(\frac{\pi}{8} + \frac{A}{2}\right) - \sin^2\left(\frac{\pi}{8} -\frac{A}{2}\right) = \frac{1}{\sqrt{2}}\sin A

  5. (tan⁡4A+tan⁡2A)(1−tan⁡23Atan⁡2A)=2tan⁡3Asec⁡2A(\tan 4A + \tan 2A)(1 - \tan^23A\tan^2A) = 2\tan 3A\sec^2A

  6. (1+tan⁡A2−sec⁡A2)(1+tan⁡A2+sec⁡A2)=sin⁡Asec⁡2A2\left(1 + \tan \frac{A}{2} - \sec\frac{A}{2}\right)\left(1 + \tan \frac{A}{2} + \sec\frac{A}{2}\right) = \sin A\sec^2\frac{A}{2}

  7. 1+sin⁡A−cos⁡A1+sin⁡A+cos⁡A=tan⁡A2\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}

  8. 1−tan⁡A21+tan⁡A2=1+sin⁡Acos⁡A=tan⁡(π4+A2)\frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}} = \frac{1 + \sin A}{\cos A} = \tan \left(\frac{\pi}{4} + \frac{A}{2}\right)

  9. cos⁡4π8+cos⁡43π8+cos⁡45π8+cos⁡47π8=32\cos^4\frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}= \frac{3}{2}

  10. 2sin⁡A−sin⁡2A2sin⁡A+sin⁡2A=tan⁡2A2\frac{2\sin A - \sin2A}{2\sin A + \sin 2A} = \tan^2\frac{A}{2}

  11. cot⁡A2−tan⁡A2=2cot⁡A\cot \frac{A}{2} - \tan \frac{A}{2} = 2\cot A

  12. 1+sin⁡A1−sin⁡A=tan⁡2(π4+A2)\frac{1 + \sin A}{1 - \sin A} = \tan^2\left(\frac{\pi}{4} + \frac{A}{2}\right)

  13. sec⁡A+tan⁡A=tan⁡(π4+A2)\sec A + \tan A = \tan\left(\frac{\pi}{4} + \frac{A}{2}\right)

  14. sin⁡A+sin⁡B−sin⁡(A+B)sin⁡A+sin⁡B+sin⁡(A+B)=tan⁡A2tan⁡B2\frac{\sin A + \sin B - \sin(A + B)}{\sin A + \sin B + \sin(A + B)} = \tan \frac{A}{2}\tan \frac{B}{2}

  15. tan⁡(π4−A2)=sec⁡A−tan⁡A=1−sin⁡A1+sin⁡A\tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \sec A - \tan A = \sqrt{\frac{1 - \sin A}{1 + \sin A}}

  16. cosec⁡(π4+A2)cosec⁡(π4−A2)=2sec⁡A\cosec\left(\frac{\pi}{4} + \frac{A}{2}\right)\cosec \left(\frac{\pi}{4} - \frac{A}{2}\right) = 2\sec A

  17. cos⁡2π8+cos⁡23π8+cos⁡25π8+cos⁡27π8=2\cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8} + \cos^2\frac{5\pi}{8} + \cos^2\frac{7\pi}{8} = 2

  18. sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8=32\sin^4\frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4\frac{5\pi}{8} + \sin^4\frac{7\pi}{8} = \frac{3}{2}

  19. (1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)=18\left(1 + \cos \frac{\pi}{8}\right)\left(1 + \cos\frac{3\pi}{8}\right)\left(1 + \cos\frac{5\pi}{8}\right)\left(1 + \cos \frac{7\pi}{8}\right) = \frac{1}{8}

  20. Find the value of sin⁡23π24\sin \frac{23\pi}{24}

  21. If A=112∘30′,A = 112^\circ30', find the value of sin⁡A\sin A and cos⁡A\cos A

Prove that

  1. sin⁡224∘−sin⁡26∘=18(5−1)\sin^224^\circ - \sin^26^\circ = \frac{1}{8}(\sqrt{5} - 1)

  2. tan⁡6∘.tan⁡42∘.tan⁡66∘.tan⁡78∘=1\tan6^\circ.\tan42^\circ.\tan66^\circ.\tan78^\circ = 1

  3. sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘=cos⁡7∘\sin47^\circ + \sin61^\circ - \sin 11^\circ - \sin25^\circ = \cos 7^\circ

  4. sin⁡12∘sin⁡48∘sin⁡54∘=18\sin 12^\circ\sin48^\circ\sin54^\circ = \frac{1}{8}

  5. cot⁡14212∘=2+3−2−6\cot 142\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}

  6. sin⁡248∘−cos⁡212∘=−5+18\sin^248^\circ - \cos^212^\circ = -\frac{\sqrt{5} + 1}{8}

  7. 4(sin⁡24∘+cos⁡6∘)=3+154(\sin 24^\circ + \cos6^\circ) = \sqrt{3} + \sqrt{15}

  8. cot⁡6∘cot⁡42∘cot⁡66∘cot⁡78∘=1\cot6^\circ\cot42^\circ\cot66^\circ\cot78^\circ = 1

  9. tan⁡12∘tan⁡24∘tan⁡48∘tan⁡84∘=1\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1

  10. sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘=116\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac{1}{16}

  11. sin⁡π5sin⁡2π5sin⁡3π5sin⁡4π5=516\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}

  12. cos⁡36∘cos⁡72∘cos⁡108∘cos⁡144∘=116\cos36^\circ\cos72^\circ\cos108^\circ\cos144^\circ = \frac{1}{16}

  13. cos⁡π15cos⁡2π15cos⁡3π15cos⁡4π15cos⁡5π15cos⁡6π15cos⁡7π15=127\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{2^7}

  14. cos⁡π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65=164\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65} = \frac{1}{64}

  15. If tan⁡A2=a−ba+btan⁡B2,\tan \frac{A}{2} = \sqrt{\frac{a - b}{a + b}}\tan \frac{B}{2}, prove that, cos⁡A=acos⁡B+ba+bcos⁡B\cos A = \frac{a\cos B + b}{a + b\cos B}

  16. If tan⁡A2 =1−e1+etan⁡B2,\tan \frac{A}{2} \ = \sqrt{\frac{1 - e}{1 + e}}\tan\frac{B}{2}, prove that cos⁡B=cos⁡A−e1−ecos⁡A\cos B = \frac{\cos A - e}{1 - e\cos A}

  17. If sin⁡A+sin⁡B=a\sin A + \sin B = a and cos⁡A+cos⁡B=b,\cos A + \cos B = b, prove that sin⁡(A+B)=2aba2+b2\sin(A + B) = \frac{2ab}{a^2 + b^2}

  18. If sin⁡A+sin⁡B=a\sin A + \sin B = a and cos⁡A+cos⁡B=b,\cos A + \cos B = b, prove that cos⁡(A−B)=12(a2+b2−2)\cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)

  19. If AA and BB be two different roots of equation acos⁡θ+bsin⁡θ=c,a\cos\theta + b\sin\theta = c, prove that

    1. tan⁡(A+B)=2aba2−b2\tan(A + B) = \frac{2ab}{a^2 - b^2}

    2. cos⁡(A+B)=a2−b2a2+b2\cos(A + B) = \frac{a^2 - b^2}{a^2 + b^2}

  20. If cos⁡A+cos⁡B=13\cos A + \cos B = \frac{1}{3} and sin⁡A+sin⁡B=14,\sin A + \sin B = \frac{1}{4}, prove that cos⁡A−B2=±524\cos \frac{A - B}{2} = \pm\frac{5}{24}

  21. If 2tan⁡A2=tan⁡B2,2\tan \frac{A}{2} = \tan \frac{B}{2}, prove that cos⁡A=3+5cos⁡B5+3cos⁡B\cos A = \frac{3 + 5\cos B}{5 + 3\cos B}

  22. If sin⁡A=45\sin A = \frac{4}{5} and cos⁡B=513,\cos B = \frac{5}{13}, prove that one value of cos⁡A−B2=865\cos \frac{A - B}{2} = \frac{8}{\sqrt{65}}

  23. If sec⁡(A+B)+sec⁡(A−B)=2sec⁡A,\sec(A + B) + \sec(A - B) = 2\sec A, prove that cos⁡B=±2cos⁡B2\cos B = \pm \sqrt{2}\cos \frac{B}{2}

  24. If cos⁡θ=cos⁡αcos⁡β1−sin⁡αsin⁡β,\cos \theta = \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}, prove that one of the values of tan⁡θ2\tan \frac{\theta}{2} is tan⁡α2−tan⁡β21−tan⁡α2tan⁡β2\frac{\tan \frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}}

  25. If tan⁡α=sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ,\tan\alpha = \frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}, prove that one of the values of tan⁡α2\tan\frac{\alpha}{2} is tan⁡θ2tan⁡ϕ2\tan\frac{\theta}{2}\tan\frac{\phi}{2}

  26. If cos⁡θ=cos⁡α+cos⁡β1+cos⁡αcos⁡β,\cos\theta = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}, prove that one of the values of tan⁡θ2\tan\frac{\theta}{2} is tan⁡α2tan⁡β2\tan\frac{\alpha}{2}\tan\frac{\beta}{2}