# TRIGONOMETRIC RATIOS OF MULTIPLE ANGLES (Part-8)

$\,1.\,$ If $\,\,\frac{\pi}{2}<\theta<\pi\,\,\text{and}\,\,\sin\theta=\frac 35,\,$ find the value of $\,\,\sin2\theta.$

Sol. $\,\,\cos\theta=-\sqrt{1-\sin^2\theta} \\ [\,\,\cos\theta<0\,\,\text{for}\,\,\frac{\pi}{2}<\theta<\pi]\\~~~~~~~~~=-\sqrt{1-(9/25)}=\sqrt{\frac{16}{25}}.\\ \therefore\,\, \sin2\theta\\=2\sin\theta\cos\theta\\=2\times \frac 35\times(-\frac 45)\\=-\frac{24}{25}.$

$\,2.\,$ If $\,\, \tan 2A=\frac 34,\,\,$ find the value of $\,\,\tan A\,\,(\frac{\pi}{2}<A<\frac{3\pi}{4})$

Sol. $\,\, \tan 2A=\frac 34 \\ \Rightarrow \frac{2\tan A}{1-\tan^2A}=\frac 34\\ \Rightarrow 8x=3-3x^2\,\,\,\,[\text{where}\,\,x=\tan A]\\ \Rightarrow 3x^2+8x-3=0 \\ \Rightarrow3x^2+9x-x-3=0 \\ \Rightarrow 3x(x+3)-1(x+3)=0 \\ \Rightarrow (x+3)(3x-1)=0\\ \Rightarrow x=-3,\,\,\frac 13 \\ \text{But, since}\,\,\frac{\pi}{2}<A<\frac{3\pi}{4},\,\,\tan A<0 \\ \text{and hence}\,\,\tan A=-3.$

$\,3.\,$ If $\,\, \cos2\theta=-\frac 12,\,\,$ find the value of $\,\,\cos\theta.$

Sol. $\,\cos2\theta=-\frac 12 \\ \Rightarrow 2\cos^2\theta-1=-\frac 12 \\ \Rightarrow 2\cos^2\theta=1-\frac 12\\~~~~~~~~~~~~~~~~~=\frac 12 \\ \Rightarrow \cos^2\theta=\frac 14 \\ \Rightarrow \cos\theta=\pm \frac 12$

$\,4.\,$ If $\,\, \sin 2A=\frac 45,\,\,$ find the value of $\,\,\sin A.(0<A<\frac{\pi}{4})$

Sol. $\,\,\sin 2A=\frac 45 \\ \therefore \,\cos 2A\\=\sqrt{1-(4/5)^2}\\=\sqrt{1-(16/25)}\\=\frac 35 \\ [\text{Since,}\,\,0<A<\frac{\pi}{4}, \,\,\,\,\cos 2A>0]\\ \therefore 1-2\sin^2A=\frac 35 \\ \Rightarrow 2\sin^2A=1-\frac 35\\ \Rightarrow \sin^2A=\frac 15 \\ \Rightarrow \sin A=\frac{1}{\sqrt5} \\ [\text{Since,}\,\,0<A<\frac{\pi}{4}, \,\,\,\,\sin A>0]$

$\,5.\,$ Prove that $\,(i)\,\,\cot \theta-\tan\theta=2\cot2\theta$

Sol. $\,(i)\,\,\cot \theta-\tan\theta\\=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\\=\frac{2(\cos^2\theta-\sin^2\theta)}{2\sin\theta\cos\theta}\\=\frac{2\cos2\theta}{\sin2\theta}\\=2\cot2\theta$

$\,5.\,$ Prove that $\,(ii)\,\,\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}-\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}=2\tan2\alpha$

Sol.  $\,(ii)\,\,\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}-\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}\\=\frac{(\cos\alpha+\sin\alpha)^2-(\cos\alpha-\sin\alpha)^2}{\cos^2\alpha-\sin^2\alpha}\\=\frac{4\cos\alpha\sin\alpha}{\cos2\alpha}\\=\frac{2\sin2\alpha}{\cos2\alpha}\\=2\tan2\alpha$

$\,5.\,$ Prove that $\,(iii)\,\,\frac{\sin3\theta}{\sin\theta}-\frac{\cos3\theta}{\cos\theta}=2$

Sol.  $\,\,\,\frac{\sin3\theta}{\sin\theta}-\frac{\cos3\theta}{\cos\theta}\\=\frac{\sin3\theta\cos\theta-\cos3\theta\sin\theta}{\sin\theta\cos\theta}\\=\frac{2\sin(3\theta-\theta)}{2\sin\theta\cos\theta}\\=\frac{2\sin2\theta}{\sin2\theta}\\=2$

$\,5.\,$ Prove that $\,(iv)\,\,\frac{\cot\theta-\tan\theta}{1-2\sin^2\theta}=\sec\theta\csc\theta$

Sol.  $\,\,\frac{\cot\theta-\tan\theta}{1-2\sin^2\theta}\\=\frac{\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}}{1-2\sin^2\theta}\\=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta\cos2\theta}\\=\frac{\cos2\theta}{\sin\theta\cos\theta\cos2\theta}\\=\csc\theta\sec\theta$

Note: Here, by csc we mean cosec .

$\,5.\,$ Prove that $\,(v)\,\,\sin^4x+\cos^4x=1-\frac 12\sin^22x$

Sol.   $\,\,\sin^4x+\cos^4x\\=(\sin^2x)^2+(\cos^2x)^2\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-2\sin^2x\cos^2x\\=1-\frac 12(2\sin x\cos x)^2\\=1-\frac 12\sin^22x$

$\,5.\,$ Prove that $\,(vi)\,\,\sin^4x+\cos^4x=\frac 12(1+2a^2-a^4),\,\,$ where $\,\,a=\sin x+\cos x$

Sol.  $\,\,\sin^4x+\cos^4x\\=(\sin^2x)^2+(\cos^2x)^2\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-2\sin^2x\cos^2x\\=-\frac 12[4\sin^2x\cos^2x-2]\\=-\frac 12[(2\sin x\cos x)^2-2]\\=-\frac 12[(1+2\sin x\cos x-1)^2-2]\\=-\frac 12[(\sin^2x+\cos^2x+2\sin x\cos x-1)^2\\-2]\\=-\frac 12[\left((\sin x+\cos x)^2-1\right)^2-2]\\=-\frac 12[(a^2-1)^2-2],\\ [\text{where},\,\,a=\sin x+\cos x]\\=-\frac 12[a^4-2a^2+1-2]\\=-\frac 12[a^4-2a^2-1]\\=\frac 12[1+2a^2-a^4]$

$\,5(vii)\,$ Prove that, $\,\,\tan A(1+\sec 2A)=\tan2A$

Sol. We have, $\,\,\tan A(1+\sec 2A)\\=\tan A \left(1+\frac{1}{\cos2A}\right)\\=\tan A\left(\frac{\cos2A+1}{\cos2A}\right)\\=\frac{\sin A}{\cos A}\times \frac{2\cos^2A}{\cos2A}\\=\frac{2\sin A\cos A}{\cos2A}\\=\frac{\sin 2A}{\cos 2A}\\=\tan 2A$

$\,5.\,$ Prove that $\,(viii)\,\,\frac{\sin\alpha+\cos\alpha}{\cos\alpha-\sin\alpha}=\tan2\alpha+\sec2\alpha$

Sol.  $\,\,\,\tan2\alpha+\sec2\alpha\\=\frac{\sin2\alpha}{\cos2\alpha}+\frac{1}{\cos2\alpha}\\=\frac{\sin2\alpha+1}{\cos2\alpha}\\=\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha-\sin^2\alpha}\\=\frac{(\sin\alpha+\cos\alpha)^2}{(\cos\alpha+\sin\alpha)(\cos\alpha-\sin\alpha)}\\=\frac{\sin\alpha+\cos\alpha}{\cos\alpha-\sin\alpha}\,\,\,\,\,\text{(proved)}$

$\,8.\,$  Evaluate :  $\,\,\,\frac{96\sin65^{\circ}\sin35^{\circ}\sin80^{\circ}}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{48(2\sin65^{\circ}\sin35^{\circ})\sin80^{\circ}}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{48(\cos30^{\circ}-\cos100^{\circ})\sin80^{\circ}}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{48(\sin80^{\circ}\cos30^{\circ}-\sin80^{\circ}\cos100^{\circ})}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{24(2\sin80^{\circ}\cos30^{\circ}-2\sin80^{\circ}\cos100^{\circ})}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{24(2\sin80^{\circ}\cos30^{\circ}-\sin180^{\circ}+\sin20^{\circ})}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=\frac{24(\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ})}{\sin20^{\circ}+2\sin80^{\circ}\cos30^{\circ}}\\=24$