Chapter 11

$$ \cos ^{2} a+\cos ^{2}\left(a+120^{\circ}\right)+\cos ^{2}\left(a-120^{\circ}\right)=\frac{3}{2} $$

$$ \cos ^{2} a+\cos ^{2}\left(a+120^{\circ}\right)+\cos ^{2}\left(a-120^{\circ}\right)=\frac{3}{2} $$

$$ (\tan 4 A+\tan 2 A)\left(1-\tan ^{2} 3 A \tan ^{2} A\right)=2 \tan 3 A \sec ^{2} A $$

$$ \frac{2 \cos 2^{n} \theta+1}{2 \cos \theta+1}=(2 \cos \theta-1)(2 \cos 2 \theta-1)\left(2 \cos 2^{2} \theta-1\right) \cdots \cdots \cdots \cdots \cdots\left(2 \cos 2^{n-1} \theta-1\right) $$

$$ \cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8}=\frac{3}{2} $$

$$ \sin ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{7 \pi}{8}=\frac{3}{2} $$

$$ \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8} $$

$$ 4 \sin 27^{\circ}=\sqrt{5+\sqrt{5}}-\sqrt{3-\sqrt{5}} $$

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

$$ \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4 $$

$$ \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=4 $$

$$ \tan 10^{\circ}+\tan 70^{\circ}-\tan 50^{\circ}=\sqrt{3} $$

$$ \tan ^{2} \frac{\pi}{16}+\tan ^{2} \frac{2 \pi}{16}+\tan ^{2} \frac{3 \pi}{16}+\tan ^{2} \frac{4 \pi}{16}+\tan ^{2} \frac{5 \pi}{16}+\tan ^{2} \frac{6 \pi}{16}+\tan ^{2} \frac{7 \pi}{16}=35 $$

$$ \cot 7 \frac{1}{2}^{\circ}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} \text { or } \tan 82 \frac{1}{2}^{\circ}=(\sqrt{2}+\sqrt{3})(1+\sqrt{2}) . $$

$$ \text { If } \cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}, \text { then show that } \tan \alpha=\sqrt{2} \tan \beta $$

$$ \text { If } \tan \theta=\frac{b}{a}, \text { prove that } a \cos 2 \theta+b \sin 2 \theta=a \text { . } $$

$$ \text { If } \tan \theta \tan \phi=\sqrt{\frac{a-b}{a+b}}, \text { prove that }(a-b \cos 2 \theta)(a-b \cos 2 \phi) \text { is independent of } \theta \text { and } \phi \text { . } $$

$$ \text { If } \tan ^{2} \theta=2 \tan ^{2} \phi+1, \text { then prove that } \cos 2 \theta+\sin ^{2} \phi=0 \text { . } $$

$$ \text { If } \tan \beta=\frac{\tan \alpha+\tan \gamma}{1+\tan \alpha \tan \gamma}, \text { prove that } \sin 2 \beta=\frac{\sin 2 \alpha+\sin 2 \gamma}{1+\sin 2 \alpha \sin 2 \gamma} \text { . } $$

$$ \sin a \sin \left(60^{\circ}-a\right) \sin \left(60^{\circ}+a\right)=\frac{1}{4} \sin 3 a $$

$$ \cos a \cos \left(60^{\circ}-a\right) \cos \left(60^{\circ}+a\right)=\frac{1}{4} \cos 3 a $$

$$ \cot a+\cot \left(60^{\circ}+a\right)-\cot \left(60^{\circ}-a\right)=3 \cot 3 a $$

$$ \cos 6 a=32 \cos ^{6} a-48 \cos ^{4} a+18 \cos ^{2} a-1 $$

$$ \cos ^{3} \theta+\cos ^{3}\left(120^{\circ}+\theta\right)+\cos ^{3}\left(240^{\circ}+\theta\right)=\frac{3}{4} \cos 3 \theta $$

$$ \text { If } \cos \theta=\frac{1}{2}\left(a+\frac{1}{a}\right), \text { then prove that } \cos 3 \theta=\frac{1}{2}\left(a^{3}+\frac{1}{a^{3}}\right) \text { . } $$

$$ (\cos \alpha+\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}=4 \cos ^{2} \frac{\alpha+\beta}{2} $$

$$ (\cos \alpha+\cos \beta)^{2}+(\sin \alpha+\sin \beta)^{2}=4 \cos ^{2} \frac{\alpha-\beta}{2} $$

$$ (\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}=4 \sin ^{2} \frac{\alpha-\beta}{2} $$

$$ \tan \left(45^{\circ}+\frac{A}{2}\right)=\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A $$

$$ \left(1+\tan \frac{\alpha}{2}-\sec \frac{\alpha}{2}\right)\left(1+\tan \frac{\alpha}{2}+\sec \frac{\alpha}{2}\right)=\sin \alpha \sec ^{2} \frac{\alpha}{2} $$

$$ \frac{1-\cos A+\cos B-\cos (A+B)}{1+\cos A-\cos B-\cos (A+B)}=\tan \frac{A}{2} \cot \frac{B}{2} $$

$$ \frac{\cos A}{1 \mp \sin A}=\tan \left(45^{\circ} \pm \frac{A}{2}\right) \text { . } $$

$$ \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}=\tan \frac{\theta}{2} $$

$$ \cot A=\frac{1}{2}\left(\cot \frac{A}{2}-\tan \frac{A}{2}\right) $$

$$ \frac{\sin (n+1) A-\sin (n-1) A}{\cos (n+1) A+2 \cos n A+\cos (n-1) A}=\tan \frac{A}{2} $$

$$ \frac{\sin (n+1) A+2 \sin n A+\sin (n-1) A}{\cos (n-1) A-\cos (n+1) A}=\cot \frac{A}{2} $$

$$ \frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}=\frac{\sqrt{3}}{2} $$

$$ 1+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}=4 \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ} $$

$$ \text { If } \cos \theta=\frac{a \cos \phi+b}{a+b \cos \phi}, \text { prove that } \tan \frac{\theta}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\phi}{2} \text { . } $$

$$ \text { If } \tan \theta=\frac{\sin \alpha \sin \beta}{\cos \alpha+\cos \beta}, \text { prove that one of the values of } \tan \frac{\theta}{2} \text { is } \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \text { . } $$

$$ \text { If } \cos A=\frac{3}{5}, \cos B=\frac{5}{13}, \text { prove that } \sin ^{2} \frac{A-B}{2}=\frac{1}{65}, \cos ^{2} \frac{A-B}{2}=\frac{64}{65} $$

$$ \text { If } \cos \theta=\frac{2 \cos \phi-1}{2-\cos \phi}, \text { prove that } \tan \frac{\theta}{2}=\sqrt{3} \tan \frac{\phi}{2}, \text { and hence show that } \sin \phi=\frac{\sqrt{3} \sin \theta}{2+\cos \theta} \text { . } $$

$$ \text { If } \sin A+\sin B=a \text { and } \cos A+\cos B=b, \text { then find the value of } \cos (A+B) \text { in terms of } a \text { and } b \text { . } $$

$$ \text { Prove that } \tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \tan \beta \tan \gamma=\frac{\sin (\alpha+\beta+\gamma)}{\cos \alpha \cos \beta \cos \gamma} \text { . } $$

$$ \text { If } A+B=45^{\circ}, \text { prove that }(\cot A-1)(\cot B-1)=2 $$

$$ \text { If } A+B=225^{\circ}, \text { prove that }\left(\frac{\cot A}{1+\cot A}\right)\left(\frac{\cot B}{1+\cot B}\right)=\frac{1}{2} \text { . } $$

If \(A+B+C=180^{\circ}\), prove that 1\. \(\sin 2 A+\sin 2 B-\sin 2 C=4 \cos A \cos B \sin C .\) ii. \(\quad \cos 2 A+\cos 2 B-\cos 2 C=1-4 \sin A \sin B \cos C .\) iii. \(\sin A+\sin B-\sin C=4 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}\). iv. \(\quad \sin ^{2} A+\sin ^{2} B-\sin ^{2} C=2 \sin A \sin B \cos C\). v. \(\quad \cos ^{2} A+\cos ^{2} B-\cos ^{2} C=1-2 \sin A \sin B \cos C\). vi. \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\). vii. \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}-\sin ^{2} \frac{C}{2}=1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\). viii. \(\sin (B+2 C)+\sin (C+2 A)+\sin (A+2 B)=4 \sin \frac{B-C}{2} \sin \frac{C-A}{2} \sin \frac{A-B}{2}\). ix. \(\quad \sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}-1=4 \sin \frac{\pi-A}{4} \sin \frac{\pi-B}{4} \sin \frac{\pi-C}{4}\).

If \(A+B+C=\frac{\pi}{2}\), prove that \(\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=1-2 \sin A \sin B \sin C\) ii. \(\quad \cos ^{2} A+\cos ^{2} B+\cos ^{2} C=2+2 \sin A \sin B \sin C\) iii. \(\cot A+\cot B+\cot C=\cot A \cot B \cot C\) iv. \(\quad \tan A \tan B+\tan B \tan C+\tan A \tan C=1\)

If \(A+B+C=2 \pi\), prove that \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}=\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\) ii. \(\quad 1-\cos ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C=0\) iii. \(\sin ^{3} A+\sin ^{3} B+\sin ^{3} C=3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}-\sin \frac{3 A}{2} \sin \frac{3 B}{2} \sin \frac{3 C}{2}\)

If \(A+B+C=2 S\), prove that 1\. \(\sin (S-A) \sin (S-B)+\sin S \sin (S-C)=\sin A \sin B\). ii. \(\quad 4 \sin S \sin (S-A) \sin (S-B) \sin (S-C)=1-\cos ^{2} A-\cos ^{2} B-\cos ^{2} C+2 \cos A \cos B \cos C .\) iii. \(\sin (S-A)+\sin (S-B)+\sin (S-C)-\sin S=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\). iv. \(\quad \cos ^{2} S+\cos ^{2}(S-A)+\cos ^{2}(S-B)+\cos ^{2}(S-C)=2+2 \cos A \cos B \cos C\). v. \(\quad \cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C=1+4 \cos S \cos (S-A) \cos (S-B) \cos (S-C)\).