# TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES (Part-1)

$\,1.\,$ If $\,\sin\alpha=\frac 45,\,\,\sin\beta=\frac35\,\,$ and $\,\alpha,\beta\,$ are positive acute angles, then find the values of $\,\,\cos\frac{\alpha-\beta}{2},\,\,\sin\frac{\alpha-\beta}{2}.$

Sol. $\,\,\cos\frac{\alpha-\beta}{2}=\sqrt{\frac12(1+\cos(\alpha-\beta))}\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}\sqrt{1+\cos\alpha\cos\beta+\sin\alpha\sin\beta}\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}\sqrt{1+\frac35.\frac45+\frac45.\frac35}\,\,[\text{By (*)}]\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}.\frac75\\~~~~~~~~~~~~~~=\frac{7\sqrt2}{10}$

Similarly, $\,\,\sin\frac{\alpha-\beta}{2}=\sqrt{\frac12(1-\cos(\alpha-\beta))}\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}\sqrt{1-\cos\alpha\cos\beta-\sin\alpha\sin\beta}\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}\sqrt{1-\frac35.\frac45-\frac45.\frac35} \,\,[\text{By (*)}]\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}.\sqrt{1-\frac{24}{25}}\\~~~~~~~~~~~~~~=\frac{1}{\sqrt2}\sqrt{\frac{1}{25}}\\~~~~~~~~~~~~~~=\frac{1}{5\sqrt2}\\~~~~~~~~~~~~~~=\frac{\sqrt2}{10}$

Note [*]: $\,\,\,\sin\alpha=\frac45 \\\Rightarrow \cos\alpha=\sqrt{1-(4/5)^2}\\~~~~~~~~~~~~~~=\frac 35 \\ \text{and}\,\,\sin\beta=\frac 35 \\ \Rightarrow \cos\beta=\sqrt{1-(3/5)^2}\\~~~~~~~~~~~~~~=\frac45.$

$\,2(i)\,$ Prove that $\,\,\frac{\cot\theta+\csc\theta}{\tan\theta+\sec\theta}=\cot\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\cot\frac{\theta}{2}$

Sol. $\,\,\frac{\cot\theta+\csc\theta}{\tan\theta+\sec\theta}\\=\frac{\cos\theta+1}{\sin\theta+1}.\frac{\cos\theta}{\sin\theta}\\=\frac{2\cos^2\frac{\theta}{2}}{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}.\frac{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}.\cot\frac{\theta}{2}\\=\frac{\cot\frac{\theta}{2}-1}{\cot\frac{\theta}{2}+1}.\cot\frac{\theta}{2}\\=\frac{\cot\frac{\pi}{4}\cot\frac{\theta}{2}-1}{\cot\frac{\theta}{2}+\cot\frac{\pi}{4}}.\cot\frac{\theta}{2}\\=\cot\left(\frac{\pi}{4}+\frac{\theta}{2}\right).\cot\frac{\theta}{2}$

Note : $\,\,\cot(A+B)=\frac{\cot A.\cot B-1}{\cot A+\cot B}$

$\,2(ii)\,$ If $\,\,\cot\frac{A}{2}-3\cot\frac{3A}{2}=\frac{4\sin A}{1+2\cos A}$

Sol. $\,\,\cot\frac{A}{2}-3\cot\frac{3A}{2}\\=\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}-3\frac{\cos\frac{3A}{2}}{\sin\frac{3A}{2}}\\=\frac{\cos\frac{A}{2}\sin\frac{3A}{2}-3\cos\frac{3A}{2}\sin\frac{A}{2}}{\sin\frac{A}{2}\sin\frac{3A}{2}}\\=\frac{2\left[\cos\frac{A}{2}\sin\frac{3A}{2}-3\cos\frac{3A}{2}\sin\frac{A}{2}\right]}{2\sin\frac{A}{2}\sin\frac{3A}{2}}\\=\frac{\sin2A+\sin A-3(\sin2A-\sin A)}{\cos A-\cos2A}\,\,[\text{By (**)}]\\=\frac{4\sin A-4\sin A\cos A}{\cos A-2\cos^2A+1}\\=\frac{4\sin A(1-\cos A)}{(1-\cos A)(1+2\cos A)}\\=\frac{4\sin A}{1+2\cos A}$

Note[**] : $\,\,\,\,\,2\sin\frac{3A}{2}\cos\frac{A}{2}\\=\sin(\frac{3A}{2}+\frac{A}{2})+\sin(\frac{3A}{2}-\frac{A}{2})\\ \,\,\,\,\, 2\cos\frac{3A}{2}\sin\frac{A}{2}\\=\sin(\frac{3A}{2}+\frac{A}{2})-\sin(\frac{3A}{2}-\frac{A}{2})$

$\,2(iii)\,$ Prove that, $\,\,\left(1+\cos{\frac{\pi}{10}}\right)\left(1+\cos{\frac{3\pi}{10}}\right)\left(1+\cos{\frac{7\pi}{10}}\right)\\ \times \left(1+\cos{\frac{9\pi}{10}}\right)=\frac{1}{16}$

Sol. $\,\,\left(1+\cos{\frac{\pi}{10}}\right)\left(1+\cos{\frac{3\pi}{10}}\right)\left(1+\cos{\frac{7\pi}{10}}\right) \\ \times \left(1+\cos{\frac{9\pi}{10}}\right)\\=\left(1+\cos{\frac{\pi}{10}}\right)\left(1+\cos{\frac{3\pi}{10}}\right)\left(1-\cos{\frac{3\pi}{10}}\right)\\ \times \left(1-\cos{\frac{\pi}{10}}\right)\\=\sin^2\frac{\pi}{10}\sin^2\frac{3\pi}{10}\\=\frac{(\sqrt5-1)^2}{16}.\frac{(\sqrt5+1)^2}{16}\\=\frac{[(\sqrt5+1)(\sqrt5-1)]^2}{16 \times 16}\\=\frac{(5-1)^2}{16\times 16}\\=\frac{1}{16}$

Note : $\,\,\cos\frac{7\pi}{10}=\cos({\pi-\frac{3\pi}{10}})=-\cos{\frac{3\pi}{10}}\\ \cos{\frac{9\pi}{10}}=\cos(\pi-\frac{\pi}{10})=-\cos{\frac{\pi}{10}}$

$\,2(iv)\,$ Prove that, $(\cos^266^{\circ}-\sin^26^{\circ})(\cos^248^{\circ}-\sin^212^{\circ})=\frac{1}{16}$

Sol. $\,\,(\cos^266^{\circ}-\sin^26^{\circ})(\cos^248^{\circ}-\sin^212^{\circ})\\=[\cos(66^{\circ}+6^{\circ})\cos(66^{\circ}-6^{\circ})]\\ \times [\cos(48^{\circ}+12^{\circ})\cos(48^{\circ}-12^{\circ})]\\=\cos72^{\circ}\cos60^{\circ}\cos60^{\circ}\cos36^{\circ}\\=\frac{\sqrt5-1}{4}.\frac 12.\frac 12.\frac{\sqrt5+1}{4}\\=\frac{(\sqrt5)^2-1^2}{4\times 2\times 2\times 4}\\=\frac{4}{4.4.4}\\=\frac{1}{16}$

$\,2(v)\,$ Prove that, $\,\,(\sin^239^{\circ}-\sin^221^{\circ})(\sin^257^{\circ}-\sin^23^{\circ})=\frac{3}{16}$

Sol. $\,\,(\sin^239^{\circ}-\sin^221^{\circ})(\sin^257^{\circ}-\sin^23^{\circ})\\=[\sin(39^{\circ}+21^{\circ})\sin(39^{\circ}-21^{\circ})] \\ \times[\sin(57^{\circ}+3^{\circ})\sin(57^{\circ}-3^{\circ})]\\=\sin60^{\circ}\sin18^{\circ}\sin60^{\circ}\sin54^{\circ}\\=\frac{\sqrt3}{2}.\frac{\sqrt5-1}{4}.\frac{\sqrt3}{2}.\frac{\sqrt5+1}{4}\\=\frac{(\sqrt3)^2 \times (\sqrt5+1)(\sqrt5-1)}{2.4.2.4}\\=\frac{3}{16}$

$\,2(vi)\,$ Prove that, $\,\,\cos^218^{\circ}\sin^236^{\circ}-\cos72^{\circ}\sin54^{\circ}\\=\frac{1}{16}$

Sol. $\,\,\cos^218^{\circ}\sin^236^{\circ}-\cos72^{\circ}\sin54^{\circ}\\=\left(\frac 14(\sqrt{10+2\sqrt5})\right)^2\left(\frac 14(\sqrt{10-2\sqrt5})\right)^2 \\-\frac{\sqrt5-1}{4}.\frac{\sqrt5+1}{4}\\=\frac{(10+2\sqrt5)(10-2\sqrt5)}{16.16}-\frac{(\sqrt5+1)(\sqrt5-1)}{4.4}\\=\frac{10^2-(2\sqrt5)^2}{16.16}-\frac{(\sqrt5)^2-1}{16}\\=\frac{80}{16 \times 16}-\frac{5-1}{16}\\=\frac{5}{16}-\frac{4}{16}\\=\frac{1}{16}$

$\,2(vii)\,$ Prove that, $\,\,\cot(7\frac 12)^{\circ}=2+\sqrt2+\sqrt3+\sqrt6$

Sol. $\,\,\cot(7\frac12)^{\circ}\\=\frac{2\cos^27\frac12^{\circ}}{2\cos7\frac12^{\circ}\sin7\frac 12^{\circ}}\\=\frac{1+\cos(2\times 7\frac12^{\circ})}{\sin(2\times 7\frac12^{\circ})}\\=\frac{1+\cos15^{\circ}}{\sin15^{\circ}}\\=\frac{1+\frac{\sqrt3+1}{2\sqrt2}}{\frac{\sqrt3-1}{2\sqrt2}}\\=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}\\=\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\\=\frac{2\sqrt6+2\sqrt2+3+\sqrt3+\sqrt3+1}{(\sqrt3)^2-1}\\=\frac{2\sqrt6+2\sqrt2+2\sqrt3+4}{2}\\=\sqrt6+\sqrt2+\sqrt3+2\\=2+\sqrt2+\sqrt3+\sqrt6.$

$\,2(viii)\,$ Prove that, $\,\,\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)\left(\cos\frac{2\pi}{10}+i\sin\frac{2\pi}{10}\right)\\ \times \left(\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right) \left(\cos\frac{4\pi}{10}+i\sin\frac{4\pi}{10}\right)=-1$

Sol. We first compute, $\,\,\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)\left(\cos\frac{4\pi}{10}+i\sin\frac{4\pi}{10}\right)\\=(\cos\alpha\cos4\alpha-\sin\alpha\sin4\alpha)\\+i(\sin\alpha\cos4\alpha+\cos\alpha\sin4\alpha)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\text{where,}\,\,\alpha=\frac{\pi}{10}]\\=\cos(\alpha+4\alpha)+i\sin(\alpha+4\alpha)\\=\cos5\alpha+i\sin5\alpha\\=\cos(\pi/2)+i\sin(\pi/2)\\=0+i.1\\=i\rightarrow(1)$

Again, $\,\left(\cos\frac{2\pi}{10}+i\sin\frac{2\pi}{10}\right)\left(\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)\\=(\cos2\alpha+i\sin2\alpha)(\cos3\alpha+i\sin3\alpha)\\=(\cos2\alpha\cos3\alpha-\sin2\alpha\sin3\alpha)\\+i(\sin2\alpha\cos3\alpha+\cos2\alpha\sin3\alpha)\\=\cos(2\alpha+3\alpha)+i\sin(2\alpha+3\alpha)\\=\cos5\alpha+i\sin5\alpha\\=\cos(\pi/2)+i\sin(\pi/2)\\=0+i.1\\=i\rightarrow(2)$

So, $\,\,\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)\left(\cos\frac{2\pi}{10}+i\sin\frac{2\pi}{10}\right)\\ \times \left(\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)\left(\cos\frac{4\pi}{10}+i\sin\frac{4\pi}{10}\right)\\=i.i\,\,\,\,[\text{By (1),(2)}]\\=-1\,\,\,\text{(proved)}$

$\,2(ix)\,$ Prove that , $\,\,\sin^4{\frac{\pi}{8}}+\sin^4{\frac{3\pi}{8}}+\sin^4{\frac{5\pi}{8}}+\sin^4{\frac{7\pi}{8}}=\frac 32$

Sol. $\,\,\sin^4{\frac{\pi}{8}}+\sin^4{\frac{3\pi}{8}}+\sin^4{\frac{5\pi}{8}}+\sin^4{\frac{7\pi}{8}}\\=\sin^4{\frac{\pi}{8}}+\sin^4{\frac{3\pi}{8}}+\sin^4\left(\pi-{\frac{3\pi}{8}}\right)+\sin^4\left({\pi-\frac{\pi}{8}}\right)\\=2\left(\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}\right)\\=2\left[\sin^4\frac{\pi}{8}+\sin^4\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right]\\=2\left[\sin^4\frac{\pi}{8}+\cos^4\frac{\pi}{8}\right]\\=2\left[(\sin^2\frac{\pi}{8})^2+(\cos^2\frac{\pi}{8})^2\right]\\=2\left[(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8})^2-2\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}\right]\\=2[1-2\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}]\\=2-(2\sin\frac{\pi}{8}\cos\frac{\pi}{8})^2\\=2-\sin^2{(2\times \frac{\pi}{8})}\\=2-\sin^2\frac{\pi}{4}\\=2-\frac 12\\=\frac 32$