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)=sinAcosB+cosAsinB\sin(A + B) = \sin A\cos B + \cos A\sin B

Substituting B=A,B = A, we get

sin2A=2sinAcosA\sin 2A = 2\sin A\cos A

Similarly, cos2A=cos2Asin2A=2cos2A1=12sin2A\cos 2A = \cos^2A - \sin^2A = 2\cos^2A - 1 = 1 - 2\sin^2A (recall formula from previous chapter and substitute B=AB = A cos2A=1sin2A\cos^2A = 1 -\sin^2A and sin2A=1cos2a\sin^2A = 1 - \cos^2a)

Also, tan2A=2tanA1tan2A\tan 2A = \frac{2\tan A}{1 - \tan^2A} (recall formula from previous chapter and put B=AB = A)

12.1.2. sin2A\sin 2A and cos2A\cos 2A in terms of tanA\tan A

sin2A=2sinAcosAsin2A+cos2A[sin2A+cos2A=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 cos2A,\cos^2A, we get

sin2A=2tanA1+tan2A\sin 2A = \frac{2\tan A}{1 + \tan^2A}

cosA=cos2Asin2A=cos2Asin2Acos2A+sin2A[sin2A+cos2A=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 cos2A,\cos^2A, we get

cos2A=1tan2A1+tan2A=cot2A1cot2A+1\cos 2A = \frac{1 - \tan^2A}{1 + \tan^2A} = \frac{\cot^2A - 1}{\cot^2A + 1}

12.1.3. Trigonometrical Ratios of 3A3A

sin3A=sin2AcosA+cos2AsinA=2sinAcos2A+cos2AsinAsin3A\sin 3A = \sin2A\cos A + \cos 2A\sin A = 2\sin A\cos^2 A + \cos^2A\sin A - \sin^3A

=2sinA(1sin2A)+(12sin2A)sinAsin3A= 2\sin A(1 - \sin^2A) + (1 - 2\sin^2A)\sin A - \sin^3A

=3sinA4sin3A= 3\sin A - 4\sin^3A

cos3A=cos2AcosAsin2AsinA=(2cos2A1)cosA2sin2AcosA\cos 3A = \cos2A\cos A - \sin 2A\sin A = (2\cos^2A - 1)\cos A - 2\sin^2 A\cos A

=2cos3AcosA2(1cos2A)cosA= 2\cos^3A - \cos A - 2(1 - \cos^2A)\cos A

=4cos3A3cosA= 4\cos^3 A - 3\cos A

We know that tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanCtanA\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

tan3A=3tanAtan3A13tan2A\tan 3A = \frac{3\tan A - \tan^3A}{1 - 3\tan^2A}

Similarly cot3A=cot3A3cotA3cot2A1\cot 3A = \frac{\cot^3 A - 3\cot A}{3\cot^2A - 1}

12.1.4. Some Important Formulae

  1. cos2A=12sin2Asin2A=12(1cos2A)\cos2A = 1 - 2\sin^2A \Rightarrow \sin^2A = \frac{1}{2}(1 - \cos2A)

  2. cos2A=2cos21cos2A=12(1+cos2A)\cos2A = 2\cos^2 - 1 \Rightarrow \cos^2A = \frac{1}{2}(1 + \cos2A)

  3. sin3A=3sinA4sin3Asin3A=12(3sinAsin3A)\sin 3A = 3\sin A - 4\sin^3A \Rightarrow sin^3A = \frac{1}{2}(3\sin A - \sin3A)

  4. cos3A=4cos3A3cosAcos3A=14(cos3A+3cosA)\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, sin2A=2sinAcosA.\sin 2A = 2\sin A\cos A. Putting A=A/2,A=A/2, we get

sinA=2sinA/2cosA/2\sin A = 2\sin A/2\cos A/2

cos2A=cos2AsinA.\cos 2A = \cos^2A - \sin^A. Putting A=A/2,A = A/2, we get

cosAcos2A2sin2A2\cos A - \cos^2\frac{A}{2} - \sin^2\frac{A}{2}

cos2A=2cos2A1.\cos 2A = 2\cos^2A - 1. Putting A=A/2,A = A/2, we get

cosA=2cos2A21\cos A = 2\cos^2\frac{A}{2} - 1

cos2A=12sin2A.\cos 2A = 1 - 2\sin^2A. Putting A=A/2,A = A/2, we get

cosA=12sin2A2\cos A = 1 - 2\sin^2\frac{A}{2}

tan2A=2tanA1tan2A.\tan 2A = \frac{2\tan A}{1 - \tan^2A}. Putting A=A/2,A = A/2, we get

tanA=2tanA21tan2A2\tan A = \frac{2\tan \frac{A}{2}}{1 - \tan^2\frac{A}{2}}

sin2A=2tanA1+tan2AsinA=2tanA21+tan2A2\sin 2A = \frac{2\tan A}{1 + \tan^2A} \therefore \sin A = \frac{2\tan \frac{A}{2}}{1 + \tan^2\frac{A}{2}}

cos2A=1tan2A1+tan2AcosA=1tan2A21+tan2A2\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}}

cot2A=cot2A12cotAcotA=cot2A212cotA2\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

sin3A=3sinA4sin3A.\sin 3A = 3\sin A - 4\sin^3A. Putting A=A3,A = \frac{A}{3}, we get

sinA=3sinA34sin3A3\sin A = \frac{3}{\sin \frac{A}{3}} - 4\sin^3\frac{A}{3}

cos3A=4cos3A3cosA\cos 3A = 4\cos^3A - 3\cos A. Putting A=A3,A = \frac{A}{3}, we get

cosA=4cos3A33cosA3\cos A = 4\cos^3\frac{A}{3} - 3\cos \frac{A}{3}

tan3A=3tanAtan3A13tan2A\tan 3A = \frac{3\tan A - \tan^3A}{1 - 3\tan^2A}

tanA=3tanA3tan3A313tan2A3\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,sinA/2cos A/2, \sin A/2 and tanA/2\tan A/2 in terms of cosA\cos A

cos2A2=1+cosA2cosA2=1+cosA2\cos^2\frac{A}{2} = \frac{1 + \cos A}{2} \therefore \cos \frac{A}{2} = \sqrt{\frac{1 + \cos A}{2}}

sin2A2=1cosA2sinA2=1cosA2\sin^2\frac{A}{2} = \frac{1 - \cos A}{}2 \therefore \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}

tan2A2=1cosA1+cosAtanA2=1cosA1+cosA\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 sinA/2\sin A/2 and cosA/2\cos A/2 in terms of sinA\sin A

(cosA2+sinA2)2=cos2A2+sin2A2+2cosA2sinA2\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+sinAcosA2+sinA2=1+sinA= 1 + \sin A \Rightarrow \cos \frac{A}{2} + \sin \frac{A}{2} = \sqrt{1 + \sin A}

Similarly, cosA2sinA2=1sinA\cos \frac{A}{2} - \sin \frac{A}{2} = \sqrt{1 - \sin A}

Adding, we get cosA2=±121+sinA±121sinA\cos \frac{A}{2} = \pm\frac{1}{2}\sqrt{1 + \sin A} \pm\frac{1}{2}\sqrt{1 - \sin A}

Subtracting, we get cosA2=±121+sinA121sinA\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 sin18\sin 18^\circ and cos72\cos 72^\circ

Let A=18,A = 18^\circ, then sin5A=902A+3A=90\sin 5A = 90^\circ \therefore 2A + 3A = 90^\circ

sin2A=sin(90sin3A)2sinAcosA=4cos3A3cosA\sin2A = \sin(90^\circ - \sin 3A) \therefore 2\sin A\cos A = 4\cos^3A - 3\cos A

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

2sinA=4cos2A3=4(1sin2A)32\sin A = 4\cos^2A - 3 = 4(1 - \sin^2A) - 3

4sin2A+2sinA1=04\sin^2A + 2\sin A - 1 = 0

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

However, since A=18sinA>0A= 18^\circ\therefore \sin A > 0

sin18=1+54\therefore \sin18^\circ = \frac{-1 + \sqrt{5}}{4}

sin(9018)=cos72=514\therefore \sin(90^\circ - 18^\circ) = \cos72^\circ = \frac{\sqrt{5} - 1}{4}

12.2.6. Value of cos18\cos 18^\circ and sin72\sin 72^\circ

cos218=1sin218=1(514)2\cos^218^\circ = 1 - \sin^218^\circ = 1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2

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

cos(9018)=sin72=1410+25\cos(90^\circ - 18^\circ) = \sin72^\circ = \frac{1}{4}\sqrt{10 + 2\sqrt{5}}

12.2.7. Value of tan18\tan 18^\circ and tan72\tan 72^\circ

tan18=sin18cos18=5110+25\tan 18^\circ = \frac{\sin18^\circ}{\cos18^\circ} = \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}

tan18cot18=1tan72=1tan18=10+2551\tan18^\circ\cot18^\circ = 1\Rightarrow \tan72^\circ = \frac{1}{\tan18^\circ} = \frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{5} - 1}

12.2.8. Value of cos36\cos 36^\circ and sin54\sin 54^\circ

cos36=12sin218=12(514)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}

sin54=sin(9036)=cos36=5+14\sin 54^\circ = \sin(90^\circ - 36^\circ) = \cos36^\circ = \frac{\sqrt{5} + 1}{4}

12.2.9. Value of sin36\sin 36^\circ and cos54\cos 54^\circ

sin36=1cos236=1(5+14)2\sin36^\circ = 1 - \cos^236^\circ = 1 - \left(\frac{\sqrt{5} + 1}{4}\right)^2

=141025= \frac{1}{4}\sqrt{10 - 2\sqrt{5}}

cos54=cos(9036)=sin36=141025\cos54^\circ = \cos(90^\circ - 36^\circ) = \sin36^\circ = \frac{1}{4}\sqrt{10 - 2\sqrt{5}}

Several other angles like, 9,15,2212,7129^\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 sin2A,\sin 2A, when

    1. cosA=35\cos A = \frac{3}{5}

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

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

  2. Find the value of cos2A,\cos 2A, when

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

    2. sinA=45\sin A = \frac{4}{5}

    3. tanA=512\tan A = \frac{5}{12}

  3. If tanA=ba,\tan A = \frac{b}{a}, find the value of acos2A+bsin2Aa\cos 2A+ b\sin 2A

Prove that

  1. sin2A1+cos2A=tanA\frac{\sin 2A}{1 + \cos 2A} = \tan A

  2. sin2A1cos2A=cotA\frac{\sin 2A}{1 - \cos 2A} = \cot A

  3. 1cos2A1+cos2A=tan2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A

  4. tanA+cotA=2cosec2A\tan A + \cot A = 2\cosec 2A

  5. tanAcotA=2cot2A\tan A - \cot A = -2\cot2A

  6. cosec2A+cot2A=cotA\cosec 2A + \cot 2A = \cot A

  7. 1cosA+cosBcos(A+B)1+cosAcosBcos(A+B)=tanA2cotB2\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. cosA1sinA=tan(45±A2)\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)

  9. sec8A1sec4A1=tan8Atan2A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}

  10. 1+tan2(45A)1tan2(45A)=cosec2A\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A

  11. sinA+sinBsinAsinB=tanA+B2tanAB2\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}}

  12. sin2Asin2BsinAcosAsinBcosB=tan(A+B)\frac{\sin^2A - \sin^2B}{\sin A\cos A - \sin B\cos B} = \tan(A + B)

  13. tan(π4+A)tan(π4A)=2tan2A\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A

  14. cosA+sinAcosAsinAcosAsinAcosA+sinA=2tan2A\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(A15)=4cos2A1+2sin2A\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}

  16. sinA+sin2A1+cosA+cos2A=tanA\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A

  17. 1+sinAcosA1+sinA+cosA=tanA2\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}

  18. sin(n+1)Asin(n1)Acos(n+1)A+2cosnA+cos(n1)A=tanA2\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+2sinnA+sin(n1)Acos(n1)cos(n+1)A=cotA2\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)AsinA=sin2(n+1)Asin2nA\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA

  21. sin(A+3B)+sin(3A+B)sin2A+sin2B=2cos(A+B)\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)

  22. sin3A+sin2AsinA=4sinAcosA2cos3A2\sin 3A + \sin 2A - \sin A = 4\sin A\cos \frac{A}{2}\cos \frac{3A}{2}

  23. tan2A=(sec2A+1)sec2A1\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}

  24. cos32A+3cos2A=4(cos6Asin6A)\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)

  25. 1+cos22A=2(cos4A+sin4A)1 + \cos^22A = 2(\cos^4A + \sin^4A)

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

  27. cosecA2cot2AcosA=2sinA\cosec A - 2\cot 2A\cos A = 2\sin A

  28. cotA=12(cotA2tanA2)\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)

  29. sinAsin(60A)sin(60+A)=14sin3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

  30. cosAcos(60A)cos(60+A)=14cos3A\cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A

  31. cotA+cot(60+A)cot(60A)=3cot3A\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A

  32. cos4A=18cos2A+8cos4A\cos 4A = 1 - 8\cos^2A + 8\cos^4A

  33. sin4A=4sinAcos3A4cosAsin3A\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A

  34. cos6A=32cos6A48cos4A+18cos2A1\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1

  35. tan3Atan2AtanA=tan3Atan2AtanA\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A

  36. 2cos2nA+12cosA+1=(2cosA1)(2cos2A1)(2cos22A1)(2cos2n11)\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 tanA=17,sinB=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(π4A)=2sec2A\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A

  2. 3cosec20sec20=4\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4

  3. tanA+2tan2A+4tan4A+8cot8A=cotA\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

  4. cos2A+cos2(2π3A)+cos2(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. 2sin2A+4cos(A+B)sinAsinB+cos2(A+B)2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B) is idnependent of A.A.

  6. If cosA=12(a+1a),\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right), show that cos2A=12(a2+1a2)\cos 2A = \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

    Prove that

  7. cos2A+sin2Acos2B=cos2B+sin2Bcos2A\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A

  8. 1+tanAtan2A=sec2A1 + \tan A\tan 2A = \sec 2A

  9. 1+sin2A1sin2A=(1+tanA1tanA)2\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2

  10. 1sin103cos10=4\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4

  11. cot2Atan2A=4cot2Acosec2A\cot^2A - \tan^2A = 4\cot2A\cosec 2A

  12. 1+sin2Acos2A=cosA+sinAcosAsinA=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. cos6Asin6A=cos2A(114sin22A)\cos^6A - \sin^6A = \cos2A\left(1 - \frac{1}{4}\sin^22A\right)

  14. cos2A+cos2(π3+A)+cos2(π3A)=32\cos^2A + \cos^2\left(\frac{\pi}{3} + A\right) + \cos^2\left(\frac{\pi}{3} - A\right)= \frac{3}{2}

  15. (1+sec2A)(1+sec22A)(1+sec23A)(1+sec2nA)=tan2nAtanA(1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA) = \frac{\tan2^nA}{\tan A}

  16. sin2nAsinA=2ncosAcos2Acos22Acos2n1A\frac{\sin2^nA}{\sin A} = 2^n\cos A\cos 2A\cos 2^2A\ldots\cos2^{n - 1}A

  17. 3(sinAcosA)4+6(sinA+cosA)2+4(sin6A+cos6A)=133(\sin A - \cos A)^4 + 6(\sin A + \cos A)^2 + 4(\sin^6A + \cos^6A) = 13

  18. 2(sin6A+cos6A)3(sin4A+cos4A)+1=02(\sin^6A + \cos^6A) - 3(\sin^4A + \cos^4A) + 1 = 0

  19. cos2A+cos2(A+B)2cosAcosBcos(A+B)\cos^2A + \cos^2(A + B) -2\cos A\cos B\cos(A + B) if independent of A.A.

  20. cos3Acos3A+sin3Asin3A=cos32A\cos^3A\cos 3A + \sin^3A\sin 3A = \cos^32A

  21. tanAtan(60A)tan(60+A)=tan3A\tan A\tan(60^\circ - A)\tan(60^\circ + A) = \tan 3A

  22. sin2A+sin3(2π3+A)+sin3(4π3+A)=34sin3A\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(cos310+sin320)=3(cos10+sin20)4(\cos^310^\circ + \sin^320^\circ) = 3(\cos 10\circ + \sin 20^\circ)

  24. sinAcos3AcosAsin3A=14sin4A\sin A\cos^3A - \cos A\sin^3A = \frac{1}{4}\sin 4A

  25. cos3Asin3A+sin3Acos3A=34sin4A\cos^3A\sin3A + \sin^3A\cos 3A = \frac{3}{4}\sin 4A

  26. sinAsin(60+A)sin(A+120)=sin3A\sin A\sin(60^\circ + A)\sin(A + 120^\circ) = \sin 3A

  27. cotA+cot(60+A)+cot(120+A)=3cot3A\cot A + \cot(60^\circ + A) + \cot(120^\circ + A) = 3\cot 3A

  28. cos5A=16cos5A20cos3A+5cosA\cos 5A = 16\cos^5A - 20\cos^3A + 5\cos A

  29. sin5A=5sinA20sin3A+16sin5A\sin 5A = 5\sin A - 20\sin^3A + 16\sin^5A

  30. cos4Acos4B=8(cosAcosB)(cosA+cosB)(cosAsinB)(cosA+sinB)\cos 4A - \cos 4B = 8(\cos A - \cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)

  31. tan4A=4tanA4tan3A16tan2A+tan4A\tan 4A = \frac{4\tan A - 4\tan^3A}{1 - 6\tan^2A + \tan^4A}

  32. If 2tanA=3tanB,2\tan A = 3\tan B, prove that tan(AB)=sin2B5cos2B\tan (A- B) = \frac{\sin 2B}{5 - \cos 2B}

  33. If sinA+sinB=x\sin A + \sin B = x and cosA+cosB=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 cosA.cos2A.cos22A..cos2n1A=12n\cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A = \frac{1}{2^n}

  35. If tanA=yx,\tan A = \frac{y}{x}, prove that xcos2A+ysin2A=xx\cos 2A + y\sin 2A = x

  36. If tan2A=1+2tan2B,\tan^2A = 1 + 2\tan^2B, prove that cos2B=1+2cos2A\cos 2B = 1 + 2\cos 2A

  37. If AA and BB lie between 00 and π2\frac{\pi}{2} and cos2A=3cos2B13cos2B,\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B}, prove that tanA=2tanB\tan A = \sqrt{2}\tan B

  38. If tanB=3tanA,\tan B = 3\tan A, prove that tan(A+B)=2sin2B1+cos2B\tan(A + B) = \frac{2\sin 2B}{1 + \cos 2B}

  39. If xsinA=ycosA,x\sin A = y\cos A, prove that xsec2A+ycosec2A=x\frac{x}{\sec 2A} + \frac{y}{\cosec 2A} = x

  40. If tanA=sec2B,\tan A = \sec 2B, prove that sin2A=1tan4B1+tan4B\sin 2A = \frac{1 - \tan^4B}{1 + \tan^4B}

  41. If A=π3,A = \frac{\pi}{3}, prove that cosA.cos2A.cos3A.cos4A.cos5A.cos6A=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 cos2A.cos4A.cos8A.cos14A=116\cos2A.\cos4A.\cos8A.\cos14A = \frac{1}{16}

  43. If tanAtanB=aba+b,\tan A\tan B = \sqrt{\frac{a - b}{a + b}}, prove that (abcos2A)(abcos2B)=a2b2(a - b\cos2A)(a - b\cos2B) = a^2 - b^2

  44. If sinA=12\sin A = \frac{1}{2} and sinB=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 cosA=1161\cos A = \frac{11}{61} and sinB=45,\sin B = \frac{4}{5}, find the value of sin2AB2\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 secA=54,\sec A = \frac{5}{4}, find tanA2\tan\frac{A}{2} and tanA.\tan A.

  47. If cosA=.3,\cos A = .3, find the value of tanA2,\tan \frac{A}{2}, and explain the resulting ambiguity.

  48. If sinA+sinB=x\sin A + \sin B = x and cosA+cosB=y,\cos A + \cos B = y, find the value of tanAB2\tan \frac{A - B}{2}

Prove that

  1. (cosA+cosB)2+(sinAsinB)2=4cos2A+B2(\cos A + \cos B)^2 + (\sin A - \sin B)^2 = 4\cos^2 \frac{A + B}{2}

  2. (cosA+cosB)2+(sinA+sinB)2=4cos2AB2(\cos A + \cos B)^2 + (\sin A + \sin B)^2 = 4\cos^2 \frac{A - B}{2}

  3. (cosAcosB)2+(sinAsinB)2=4sin2AB2(\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 4\sin^2 \frac{A - B}{2}

  4. sin2(π8+A2)sin2(π8A2)=12sinA\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. (tan4A+tan2A)(1tan23Atan2A)=2tan3Asec2A(\tan 4A + \tan 2A)(1 - \tan^23A\tan^2A) = 2\tan 3A\sec^2A

  6. (1+tanA2secA2)(1+tanA2+secA2)=sinAsec2A2\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+sinAcosA1+sinA+cosA=tanA2\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}

  8. 1tanA21+tanA2=1+sinAcosA=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. cos4π8+cos43π8+cos45π8+cos47π8=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. 2sinAsin2A2sinA+sin2A=tan2A2\frac{2\sin A - \sin2A}{2\sin A + \sin 2A} = \tan^2\frac{A}{2}

  11. cotA2tanA2=2cotA\cot \frac{A}{2} - \tan \frac{A}{2} = 2\cot A

  12. 1+sinA1sinA=tan2(π4+A2)\frac{1 + \sin A}{1 - \sin A} = \tan^2\left(\frac{\pi}{4} + \frac{A}{2}\right)

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

  14. sinA+sinBsin(A+B)sinA+sinB+sin(A+B)=tanA2tanB2\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(π4A2)=secAtanA=1sinA1+sinA\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(π4A2)=2secA\cosec\left(\frac{\pi}{4} + \frac{A}{2}\right)\cosec \left(\frac{\pi}{4} - \frac{A}{2}\right) = 2\sec A

  17. cos2π8+cos23π8+cos25π8+cos27π8=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. sin4π8+sin43π8+sin45π8+sin47π8=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+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)=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 sin23π24\sin \frac{23\pi}{24}

  21. If A=11230,A = 112^\circ30', find the value of sinA\sin A and cosA\cos A

Prove that

  1. sin224sin26=18(51)\sin^224^\circ - \sin^26^\circ = \frac{1}{8}(\sqrt{5} - 1)

  2. tan6.tan42.tan66.tan78=1\tan6^\circ.\tan42^\circ.\tan66^\circ.\tan78^\circ = 1

  3. sin47+sin61sin11sin25=cos7\sin47^\circ + \sin61^\circ - \sin 11^\circ - \sin25^\circ = \cos 7^\circ

  4. sin12sin48sin54=18\sin 12^\circ\sin48^\circ\sin54^\circ = \frac{1}{8}

  5. cot14212=2+326\cot 142\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}

  6. sin248cos212=5+18\sin^248^\circ - \cos^212^\circ = -\frac{\sqrt{5} + 1}{8}

  7. 4(sin24+cos6)=3+154(\sin 24^\circ + \cos6^\circ) = \sqrt{3} + \sqrt{15}

  8. cot6cot42cot66cot78=1\cot6^\circ\cot42^\circ\cot66^\circ\cot78^\circ = 1

  9. tan12tan24tan48tan84=1\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1

  10. sin6sin42sin66sin78=116\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac{1}{16}

  11. sinπ5sin2π5sin3π5sin4π5=516\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}

  12. cos36cos72cos108cos144=116\cos36^\circ\cos72^\circ\cos108^\circ\cos144^\circ = \frac{1}{16}

  13. cosπ15cos2π15cos3π15cos4π15cos5π15cos6π15cos7π15=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. cosπ65cos2π65cos4π65cos8π65cos16π65cos32π65=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 tanA2=aba+btanB2,\tan \frac{A}{2} = \sqrt{\frac{a - b}{a + b}}\tan \frac{B}{2}, prove that, cosA=acosB+ba+bcosB\cos A = \frac{a\cos B + b}{a + b\cos B}

  16. If tanA2 =1e1+etanB2,\tan \frac{A}{2} \ = \sqrt{\frac{1 - e}{1 + e}}\tan\frac{B}{2}, prove that cosB=cosAe1ecosA\cos B = \frac{\cos A - e}{1 - e\cos A}

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

  18. If sinA+sinB=a\sin A + \sin B = a and cosA+cosB=b,\cos A + \cos B = b, prove that cos(AB)=12(a2+b22)\cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)

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

    1. tan(A+B)=2aba2b2\tan(A + B) = \frac{2ab}{a^2 - b^2}

    2. cos(A+B)=a2b2a2+b2\cos(A + B) = \frac{a^2 - b^2}{a^2 + b^2}

  20. If cosA+cosB=13\cos A + \cos B = \frac{1}{3} and sinA+sinB=14,\sin A + \sin B = \frac{1}{4}, prove that cosAB2=±524\cos \frac{A - B}{2} = \pm\frac{5}{24}

  21. If 2tanA2=tanB2,2\tan \frac{A}{2} = \tan \frac{B}{2}, prove that cosA=3+5cosB5+3cosB\cos A = \frac{3 + 5\cos B}{5 + 3\cos B}

  22. If sinA=45\sin A = \frac{4}{5} and cosB=513,\cos B = \frac{5}{13}, prove that one value of cosAB2=865\cos \frac{A - B}{2} = \frac{8}{\sqrt{65}}

  23. If sec(A+B)+sec(AB)=2secA,\sec(A + B) + \sec(A - B) = 2\sec A, prove that cosB=±2cosB2\cos B = \pm \sqrt{2}\cos \frac{B}{2}

  24. If cosθ=cosαcosβ1sinαsinβ,\cos \theta = \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta},