Differentiation (Part-38) | S N De

Unit-2 : TRIGONOMETRY

1. Trigonometric Functions: 

Positive and negative angles. Measuring angles in radians and in degrees and conversion from one measure to another. Definition of trigonometric functions with the help of unit circle. Truth of the identity $\,\sin^2 x+\cos^2 x = 1,\forall x\,\,$. Signs of trigonometric functions and sketch of their graphs. Expressing $\,\,\sin(x + y)\,$ and $\,\,\cos(x+y)\,\,$ in terms of $\sin x, \sin y, \cos x \,\,\text{and}\,\, \cos y.$ Deducing the identities like the following: $$\tan(x \pm y)=\frac{\tan x \pm \tan y}{1 \mp \tan x.\tan y},\\ \cot(x \pm y)=\frac{\cot x\cot y \mp 1}{\cot y \pm \cot x} \\ \sin x +\sin y=2\sin \frac{x+y}{2}. \cos\frac{x-y}{2},\\ \cos x+\cos y=2\cos \frac{x+y}{2}. \cos\frac{x-y}{2} \\ \sin x -\sin y=2\cos \frac{x+y}{2}. \sin\frac{x-y}{2},\\ \cos x -\cos y=-2\sin \frac{x+y}{2}. \sin\frac{x-y}{2}$$

Identities related to $\,\,\sin 2x,\,\cos 2x,\,\tan 2x,\,\sin 3x\,\,\text{and}\,\,\tan 3x.\,\,$ General solution of trigonometric equations of the type $\,\,\sin \theta=\sin \alpha,\cos \theta=\cos \alpha\,\,\text{and}\,\,\tan \theta=\tan \alpha.\,\,$ Proof and simple application of sine and cosine rules only.

 1. Prove that , $\,(i)\, \cot(\theta+15^{\circ})-\tan(\theta-15^{\circ})=\frac{4 \cos {2\theta}}{1+2\sin{2\theta}}$

Sol. $\,(i)\, \cot(\theta+15^{\circ})-\tan(\theta-15^{\circ})\\=\frac{\cos(\theta+15^{\circ})}{\sin(\theta+15^{\circ})}-\frac{\sin(\theta-15^{\circ})}{\cos(\theta-15^{\circ})}\\=\frac{\cos(\theta+15^{\circ})\cos(\theta-15^{\circ})-\sin(\theta+15^{\circ})\sin(\theta-15^{\circ})}{\sin(\theta+15^{\circ})\cos(\theta-15^{\circ})}\\=\frac{\cos(\theta+15^{\circ}+\theta-15^{\circ})}{\frac 12(\sin{2 \theta}+\sin{30^{\circ}})}\\=\frac{4\cos{2\theta}}{1+2\sin{2\theta}}\,\,\,[\text{Using,}\,\,\sin{30^{\circ}}=\frac 12]$

1. Prove that , $\,(ii)\, \,\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)=2\sec{2 \theta}$

Sol. $\,\, \tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)\\=\frac{\sin \left(\frac{\pi}{4}+\theta\right)}{\cos\left(\frac{\pi}{4}+\theta\right)}+\frac{\sin\left(\frac{\pi}{4}-\theta\right)}{\cos\left(\frac{\pi}{4}-\theta\right)}\\= \frac{\sin\left(\frac{\pi}{4}+\theta\right)\cos\left(\frac{\pi}{4}-\theta\right)+\cos\left(\frac{\pi}{4}+\theta\right)\sin\left(\frac{\pi}{4}-\theta\right)}{\cos \left(\frac{\pi}{4}+\theta\right)\cos\left(\frac{\pi}{4}-\theta\right)}\\=\frac{\sin\left(\frac{\pi}{4}+\theta+\frac{\pi}{4}-\theta\right)}{ \frac 12\left(\cos\frac{\pi}{2}+\cos{2\theta}\right)}\\=2 \sec{2 \theta}$

$(iii)\,\, \cos^3{\theta}\cos{3\theta}+\sin^3\theta\sin{3\theta}=\cos^3{2\theta}$

Sol. $\,\, \cos^3{\theta}\cos{3\theta}+\sin^3\theta\sin{3\theta}\\=\frac{3\cos\theta+\cos{3\theta}}{4}.\cos{3\theta}+\frac{3\sin \theta-\sin{3\theta}}{4}.\sin{3\theta}\\=\frac 14[3(\cos \theta\cos{3\theta}+\sin \theta \sin{3\theta})\\+(\cos^2{3\theta}-\sin^2{3\theta})]\\=\frac 14(3 \cos (3\theta-\theta)+\cos 2.3\theta)\\=\frac 14(3 \cos 2\theta+\cos 2.3\theta)\\=\cos^32\theta$

$(iv)\,\,\cos^3\alpha+\cos^3(120^{\circ}+\alpha)+\cos^3(240^{\circ}+\alpha)=\frac 34 \cos 3\alpha$

Sol. $\,\,\cos^3\alpha+\cos^3(120^{\circ}+\alpha)+\cos^3(240^{\circ}+\alpha)\\=\frac 14 [\cos 3 \alpha+3\cos\alpha+\cos(360^{\circ}+3\alpha)\\+3\cos(120^{\circ}+\alpha)+\cos(720^{\circ}+3\alpha)\\+3\cos(240^{\circ}+\alpha)]\\=\frac 14[3\cos3\alpha+3\cos\alpha+3\{\cos(120^{\circ}+\alpha)\\+\cos(240^{\circ}+\alpha)\}]\\=\frac 14[3\cos 3\alpha+3\cos \alpha+3.2\cos(180^{\circ}+\alpha) \\ \times \cos60^{\circ}]\\=\frac 34\cos 3\alpha$

Additional Notes : $\,4\cos^3(120^{\circ}+\alpha)=\cos [3(120^{\circ}+\alpha)]\\+3 \cos(120^{\circ}+\alpha),\\ 4\cos^3(240^{\circ}+\alpha)=\cos [3(240^{\circ}+\alpha)]\\+3 \cos(240^{\circ}+\alpha), \\ \cos 3 \alpha=\cos(360^{\circ}+3\alpha)=\cos(720^{\circ}+3\alpha)$

$\,(v)\, \tan 70^{\circ}-\tan 50^{\circ}+\tan 10^{\circ}=\sqrt3$

Sol. $\,\tan 70^{\circ}-\tan 50^{\circ}+\tan 10^{\circ}\\=\tan (60^{\circ}+10^{\circ})-\tan(60^{\circ}-10^{\circ})+\tan 10^{\circ}\\=\frac{\tan 60^{\circ}+\tan 10^{\circ}}{1-\tan 60^{\circ} \tan 10^{\circ}}-\frac{\tan 60^{\circ}-\tan 10^{\circ}}{1+\tan 60^{\circ}\tan 10^{\circ}}+\tan 10^{\circ}\\=\frac{\sqrt3+\tan 10^{\circ}}{1-\sqrt3\tan 10^{\circ}}-\frac{\sqrt3-\tan 10^{\circ}}{1+\sqrt3\tan 10^{\circ}}+\tan 10^{\circ}\\=\frac{\sqrt3+3\tan 10^{\circ}+\tan 10^{\circ}+\sqrt3 \tan^210^{\circ}-(\sqrt3-3\tan 10^{\circ}\\-\tan 10^{\circ}+\sqrt3 \tan^210^{\circ})}{1-3\tan^210^{\circ}}\\+\tan 10^{\circ}\\=\frac{8\tan 10^{\circ}}{1-3\tan^210^{\circ}}+\tan 10^{\circ}\\=\frac{9 \tan 10^{\circ}-3\tan^3 10^{\circ}}{1-3\tan^210^{\circ}} \\=3\left( \frac{3\tan 10^{\circ}-3\tan^310^{\circ}}{1-3\tan^210^{\circ}}\right)\\=3\tan30^{\circ}\\=\sqrt3$

$\,(vi)\, \tan \theta \tan(\frac{\pi}{3}+\theta)\tan(\frac{\pi}{3}-\theta)=\tan 3\theta$

Sol. $\,\, \tan \theta \tan(\frac{\pi}{3}+\theta)\tan(\frac{\pi}{3}-\theta)\\=\tan \theta . \frac{\sqrt3+\tan \theta}{1-\sqrt3 \tan \theta}. \frac{\sqrt3-\tan \theta}{1+\sqrt3 \tan \theta}\\=\frac{3\tan \theta-\tan^3\theta}{1-3\tan^2\theta}\\=\tan \theta$

$\,(vii)\, \tan \theta+2\tan 2\theta+4 \tan 4\theta+8\cot 8\theta=\cot \theta$

Sol. $\, \tan \theta+2\tan 2\theta+4 \tan 4\theta+8\cot 8\theta\\=\tan \theta+2\tan 2\theta+4 \tan 4\theta+\frac{8(1-\tan^2 4\theta)}{2\tan 4\theta}\\=\tan \theta+2\tan 2\theta+\frac{8\tan^24\theta+8-8\tan^24\theta}{2\tan 2\theta}\\=\tan \theta+2\tan 2\theta+\frac{4}{\tan 2\theta}\\=\tan \theta+2\tan 2\theta +\frac{4(1-\tan^2 2\theta)}{2\tan 2\theta}\\=\tan \theta+\frac{4 \tan^2 2\theta+4-4\tan^2 2\theta}{2\tan 2\theta}\\=\tan \theta+\frac{2}{\tan 2\theta}\\=\tan \theta+\frac{2(1-\tan^2\theta)}{2\tan \theta}\\=\cot \theta$

$\,(viii)\, \cos \frac{\pi}{11}\cos \frac{2\pi}{11}\cos \frac{3\pi}{11}\cos \frac{4\pi}{11}\cos \frac{5\pi}{11} =\frac{1}{32}$

Sol.  Let $\,\,\frac{\pi}{11}=\theta  \Rightarrow \theta=11 \pi \\ \therefore P=\cos \theta \cos 2\theta \cos 3\theta \cos 4\theta \cos 5\theta \,\,,\text{say}\\ \Rightarrow 2P\sin \theta=(2 \sin  \theta\cos \theta) \cos 2 \theta \cos3 \theta \cos 4 \theta\cos 5 \theta \\ \Rightarrow 2P\sin  \theta=\sin 2 \theta\cos 2 \theta\cos3 \theta\cos4 \theta\cos5 \theta \\ \Rightarrow 2^2P\sin  \theta=2\sin 2 \theta\cos 2 \theta\cos3 \theta\cos4 \theta\cos5 \theta \\ \Rightarrow 2^2P \sin \theta=\sin 4\theta \cos 4\theta \cos 3\theta \cos 5\theta \\ \Rightarrow 2^3P \sin \theta=(2\sin 4\theta \cos 4\theta )\cos 3\theta \cos 5\theta \\ \Rightarrow 2^3 P\sin \theta=\sin 8\theta \cos 3\theta\cos 5\theta \\~~~~~~~~~~~~~~~~~~~~=\sin(11 \theta-3\theta)\cos 3\theta \cos 5\theta\\~~~~~~~~~~~~~~~~~~~~=\sin(\pi-3\theta)\cos 3\theta\cos 5\theta\\~~~~~~~~~~~~~~~~~~~~=\sin 3\theta \cos 3\theta\cos 5\theta \\ \Rightarrow 2^4P\sin\theta=(2\sin 3\theta\cos 3\theta)\cos 5\theta\\ \Rightarrow 2^4P\sin \theta=\sin 6\theta\cos 5\theta \\~~~~~~~~~~~~~~~~~~~~=\sin(11 \theta-5\theta)\cos 5\theta\\~~~~~~~~~~~~~~~~~~~~=\sin(\pi-5\theta)\cos 5\theta\\~~~~~~~~~~~~~~~~~~~~=\sin 5\theta\cos5\theta\\ \Rightarrow2^5P\sin\theta=2\sin 5\theta\cos 5\theta \\ \Rightarrow 2^5P\sin\theta=\sin (2 \times 5\theta)\\ \Rightarrow 2^5P \sin\theta=\sin 10\theta \\ \Rightarrow 2^5P \sin \theta=\sin(11\theta-\theta)\\ \Rightarrow 2^5P\sin\theta=\sin(\pi-\theta)\\ \Rightarrow 2^5P \sin\theta=\sin \theta \\ \Rightarrow P=\frac{1}{2^5} =\frac{1}{32}$

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