Differentiation (Part-38) | S N De

Problem 1(xxxviii)

$$ 3\tan\left(\theta-\frac{\pi}{12}\right) = \tan\left(\theta+\frac{\pi}{12}\right) \quad \text{--- (1)} $$

Let $\theta-\frac{\pi}{12}=p$ and $\theta+\frac{\pi}{12}=q$.
So that, $q+p=2\theta \quad \text{--- (2)}$, and $q-p=\frac{2\pi}{12}=\frac{\pi}{6} \quad \text{--- (3)}$.

Then, from (1) we get:

$$ 3\tan p = \tan q \\ \Rightarrow 3\frac{\sin p}{\cos p} = \frac{\sin q}{\cos q} \\ \Rightarrow 3 = \frac{\sin q \cos p}{\cos q \sin p} \\ \Rightarrow \frac{3+1}{3-1} = \frac{\sin q\cos p+\cos q \sin p}{\sin q\cos p-\cos q\sin p} \\ \Rightarrow \frac{\sin (q+p)}{\sin(q-p)} = \frac{4}{2} \\ \Rightarrow \frac{\sin 2\theta}{\sin (\pi/6)} = 2 \quad [\text{By (2), (3)}] \\ \Rightarrow \sin 2\theta = 2\sin(\pi/6) \\ \Rightarrow \sin 2\theta = 2 \times \frac{1}{2} = 1 \\ \Rightarrow 2\theta = (4n+1)\frac{\pi}{2}, \quad n\in\mathbb{Z} \\ \Rightarrow \theta = \frac{(4n+1)\pi}{4} = n\pi + \frac{\pi}{4}, \quad n\in\mathbb{Z} \quad \text{(ans.)} $$

Problem 1(xxxix)

$$ \sec4\theta-\sec2\theta=2 \\ \Rightarrow \frac{1}{\cos4\theta}-\frac{1}{\cos2\theta}=2 \\ \Rightarrow \frac{\cos2\theta-\cos4\theta}{\cos4\theta\cos2\theta}=2 \\ \Rightarrow \cos2\theta-\cos4\theta=2\cos4\theta\cos2\theta \\ \Rightarrow \cos2\theta-\cos4\theta=\cos6\theta+\cos2\theta \\ \Rightarrow \cos6\theta+\cos4\theta=0 \\ \Rightarrow 2\cos\left(\frac{6\theta+4\theta}{2}\right) \cos\left(\frac{6\theta-4\theta}{2}\right)=0 \\ \Rightarrow 2\cos5\theta\cos\theta=0 \\ \therefore \cos5\theta=0 \quad \text{--- (1)}, \\ \text{or,} \quad \cos\theta=0 \quad \text{--- (2)} $$

From (1), we get:

$$ 5\theta = (2n+1)\frac{\pi}{2}, \quad n\in\mathbb{Z} \\ \Rightarrow \theta = \frac{(2n+1)\pi}{10} \quad \text{--- (3)} $$

Again, by (2), we get:

$$ \theta = (2n+1)\frac{\pi}{2} \\ \Rightarrow \theta = \frac{(10n+5)\pi}{10} \\ \Rightarrow \theta = [2(5n+2)+1]\frac{\pi}{10} \quad \text{--- (4)} $$

Hence, from (3) and (4), it follows that the general solution is given by:

$$ \theta = \frac{(2n+1)\pi}{10}, \quad n\in\mathbb{Z} $$

Problem 1(xL)

$$ \sin x+\cos x=\frac{\cos 2x}{1-\sin2x} \\ \Rightarrow \sin x+\cos x=\frac{\cos^2x-\sin^2x}{\cos^2x-2\sin x\cos x+\sin^2x} \\ \Rightarrow \sin x+\cos x=\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x-\sin x)^2} \\ \Rightarrow \sin x+\cos x=\frac{\cos x+\sin x}{\cos x-\sin x} \\ \Rightarrow (\sin x+\cos x)(\cos x-\sin x) = (\cos x+\sin x) \\ \Rightarrow (\sin x+\cos x)(\cos x-\sin x-1)=0 \\ \therefore \sin x+\cos x=0 \quad \text{--- (1)}, \\ \text{or,} \quad \cos x-\sin x-1=0 \quad \text{--- (2)} $$

From (1), we get:

$$ \sin x = -\cos x \\ \Rightarrow \tan x = -1 \\ \Rightarrow x = n\pi - \frac{\pi}{4}, \quad n\in\mathbb{Z} $$

Again, by (2), we get:

$$ \frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{\sqrt{2}} \\ \Rightarrow \cos\left(x+\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \\ \Rightarrow x+\frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} \\ \Rightarrow x = 2n\pi, \quad 2n\pi - \frac{\pi}{2} $$

Hence, the required general solution is given by:

$$ x = n\pi - \frac{\pi}{4}, \quad 2n\pi, \quad 2n\pi - \frac{\pi}{2}, \quad n\in\mathbb{Z} $$

Problem 2

Find the general value of $\theta$ which satisfies $2\cos\theta+1=0$ and $2\sin\theta-\sqrt{3}=0$.

Solution:

$$ 2\cos\theta+1=0 \\ \Rightarrow \cos\theta = -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right) \\ \Rightarrow \theta = 2n\pi \pm \frac{2\pi}{3} \quad \text{--- (1)} $$

Again,

$$ 2\sin\theta-\sqrt{3}=0 \\ \Rightarrow \sin\theta = \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \\ \Rightarrow \theta = n\pi + (-1)^n \frac{\pi}{3} \quad \text{--- (2)} $$

Now, for $n$ even $= 2k$ (say), we have from (2):

$$ \theta = 2k\pi + \frac{\pi}{3} \quad \text{--- (3)} $$

Combining (1) and (3), we get the general solution:

$$ \theta = 2n\pi + \frac{2\pi}{3}, \quad n\in\mathbb{Z} $$

Problem 3

If $\cos x+\sin x=\cos\alpha-\sin\alpha$, prove that $x-\frac{\pi}{4}=2n\pi \pm \left(\alpha+\frac{\pi}{4}\right)$.

Solution:

$$ \cos x+\sin x=\cos \alpha-\sin\alpha \\ \Rightarrow \cos x-\cos\alpha = -(\sin x+\sin\alpha) \\ \Rightarrow -2 \sin\left(\frac{x+\alpha}{2}\right)\sin\left(\frac{x-\alpha}{2}\right) = -2\sin\left(\frac{x+\alpha}{2}\right)\cos\left(\frac{x-\alpha}{2}\right) \\ \Rightarrow \sin\left(\frac{x+\alpha}{2}\right)\left[\sin\left(\frac{x-\alpha}{2}\right)-\cos\left(\frac{x-\alpha}{2}\right)\right]=0 \\ \therefore \sin\left(\frac{x+\alpha}{2}\right)=0 \quad \text{--- (1)}, \\ \text{or,} \quad \sin\left(\frac{x-\alpha}{2}\right)-\cos\left(\frac{x-\alpha}{2}\right)=0 \quad \text{--- (2)} $$

From (1), we get:

$$ \frac{x+\alpha}{2} = n\pi \\ \Rightarrow x = 2n\pi - \alpha \\ \Rightarrow x - \frac{\pi}{4} = 2n\pi - \left(\alpha + \frac{\pi}{4}\right) \quad \text{--- (3)} $$

Again, by (2), we get:

$$ \sin\left(\frac{x-\alpha}{2}\right) = \cos\left(\frac{x-\alpha}{2}\right) \\ \Rightarrow \tan\left(\frac{x-\alpha}{2}\right) = 1 \\ \Rightarrow \frac{x-\alpha}{2} = n\pi + \frac{\pi}{4} \\ \Rightarrow x - \alpha = 2n\pi + \frac{\pi}{2} \\ \Rightarrow x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2} - \frac{\pi}{4} + \alpha \\ \Rightarrow x - \frac{\pi}{4} = 2n\pi + \left(\alpha + \frac{\pi}{4}\right) \quad \text{--- (4)} $$

Hence, from (3) and (4), we get:

$$ x - \frac{\pi}{4} = 2n\pi \pm \left(\alpha + \frac{\pi}{4}\right) \quad \text{(Proved)} $$

Alternative Method:

$$ \cos x+\sin x = \cos\alpha-\sin\alpha \\ \Rightarrow \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}}\cos\alpha-\frac{1}{\sqrt{2}}\sin\alpha \\ \Rightarrow \cos\left(x-\frac{\pi}{4}\right) = \cos\left(\alpha+\frac{\pi}{4}\right) \\ \therefore x-\frac{\pi}{4} = 2n\pi \pm \left(\alpha+\frac{\pi}{4}\right), \quad n\in\mathbb{Z} \quad \text{(Proved)} $$

Problem 4

If $\tan ax=\tan bx \ (a \neq b)$, show that the values of $x$ form an A.P.

Solution:

$$ \tan ax = \tan bx \\ \Rightarrow ax = n\pi + bx \\ \Rightarrow x(a-b) = n\pi \\ \Rightarrow x = \frac{n\pi}{a-b}, \quad n\in\mathbb{Z} \quad \text{--- (1)} $$

Hence, from (1) we get, the values of $x$ are given by:

$$ \cdots, \frac{-3\pi}{a-b}, \frac{-2\pi}{a-b}, \frac{-\pi}{a-b}, 0, \frac{\pi}{a-b}, \frac{2\pi}{a-b}, \frac{3\pi}{a-b}, \cdots $$

which clearly are in A.P. with common difference $\frac{\pi}{a-b}$.

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Read More Topics

• COMPLEX NUMBERS
• QUADRATIC EQUATIONS
• TRIGONOMETRIC RATIOS OF MULTIPLE ANGLES
• RELATION AND MAPPING
• TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES
• TRIGONOMETRIC RATIOS OF COMPOUND ANGLES