Differentiation (Part-38) | S N De

Welcome to Part 2 of our comprehensive solution series for Transformations of Sums and Products. In this article, we will solve problems 6 through 11 from the Chhaya Publication textbook for Class XI. These solutions cover key trigonometric identities and transformations essential for WBCHSE and other board exams.

Solutions: Problems 6-11

6.

If $A+B+C=\pi$, $\sin(A+\frac C2)=n\sin\frac C2$, show that $\tan\frac A2\tan \frac B2=\frac{n-1}{n+1}$.

Solution

We have:

$$\sin\left(A+\frac C2\right)=n\sin\frac C2 \quad \cdots(1)$$ $$A+B+C=\pi \quad \cdots(2)$$

Now consider the RHS:

$$\frac{n-1}{n+1} = \frac{\frac{\sin(A+C/2)}{\sin(C/2)}-1}{\frac{\sin(A+C/2)}{\sin(C/2)}+1}$$ $$= \frac{\sin(A+C/2)-\sin(C/2)}{\sin(A+C/2)+\sin(C/2)}$$ $$= \frac{2\sin(A/2)\cos((A+C)/2)}{2\sin((A+C)/2)\cos(A/2)}$$ $$= \tan\frac A2\cot\left(\frac{\pi}{2}-\frac B2\right) \quad [\text{By (2)}]$$ $$\therefore \tan\frac A2\tan\frac B2=\frac{n-1}{n+1}$$

7.

If $\sin\theta+\sin\phi=a$, $\cos\theta+\cos\phi=b$, prove that $\tan\frac{\theta-\phi}{2}=\pm\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$.

Solution

Using sum-to-product formulas:

$$a=\sin\theta+\sin\phi=2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$$ $$b=\cos\theta+\cos\phi=2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$$

Squaring and adding:

$$a^2+b^2 = 4\cos^2\frac{\theta-\phi}{2}\left(\sin^2\frac{\theta+\phi}{2}+\cos^2\frac{\theta+\phi}{2}\right)$$ $$= 4\cos^2\frac{\theta-\phi}{2}$$

Therefore:

$$\tan^2\frac{\theta-\phi}{2} = \frac{1-\cos^2\frac{\theta-\phi}{2}}{\cos^2\frac{\theta-\phi}{2}} = \frac{1-\frac{a^2+b^2}{4}}{\frac{a^2+b^2}{4}} = \frac{4-a^2-b^2}{a^2+b^2}$$

8.

Prove that $\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n=2\cot^n\frac{A-B}{2}$ or $0$ accordingly as $n$ is even or odd.

Solution

Let $P$ be the given expression. Using sum-to-product formulas:

$$P=\left(\frac{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}{2\sin\frac{A-B}{2}\cos\frac{A+B}{2}}\right)^n+\left(\frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{-2\sin\frac{A-B}{2}\sin\frac{A+B}{2}}\right)^n$$ $$= \left[\cot\frac{A-B}{2}\right]^n + \left[-\cot\frac{A-B}{2}\right]^n \quad \cdots(1)$$

From (1):

• If $n$ is even: $P = \cot^n\frac{A-B}{2} + \cot^n\frac{A-B}{2} = 2\cot^n\frac{A-B}{2}$
• If $n$ is odd: $P = \cot^n\frac{A-B}{2} - \cot^n\frac{A-B}{2} = 0$

9.

If $\sin A=\frac{1}{\sqrt2}$, $\sin B=\frac{1}{\sqrt3}$, find $\tan\frac{A+B}{2}\cot\frac{A-B}{2}$.

Solution

$$\tan\frac{A+B}{2}\cot\frac{A-B}{2} = \frac{\sin A+\sin B}{\sin A-\sin B}$$ $$= \frac{\frac{1}{\sqrt2}+\frac{1}{\sqrt3}}{\frac{1}{\sqrt2}-\frac{1}{\sqrt3}} = \frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}$$ $$= \frac{(\sqrt3+\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2} = \frac{3+2+2\sqrt6}{1} = 5+2\sqrt6$$

10.

Find the value of $A$ for which $\cos A\sin (A-\frac{\pi}{6})$ is maximum.

Solution

$$\cos A\sin\left(A-\frac{\pi}{6}\right) = \frac 12\left[2\cos A\sin\left(A-\frac{\pi}{6}\right)\right]$$ $$= \frac 12\left[\sin\left(2A-\frac{\pi}{6}\right)-\sin\frac{\pi}{6}\right] \quad \cdots(1)$$

This is maximum when $\sin(2A-\pi/6)=1$:

$$2A-\frac{\pi}{6}=\frac{\pi}{2} \implies A=\frac{\pi}{3}$$

11.

The two different values $\theta_1$ and $\theta_2$ ($0 \leq \theta < 2\pi$) satisfy $\sin(\theta+\phi)=\frac 12\sin2\phi$. Prove that $\frac{\sin\theta_1+\sin \theta_2}{\cos\theta_1+\cos \theta_2}=\cot \phi$.

Solution

Since both values satisfy the equation:

$$\sin(\theta_1+\phi)=\sin(\theta_2+\phi)$$ $$\Rightarrow \theta_1+\phi=\pi-(\theta_2+\phi) \quad [\theta_1 \neq \theta_2]$$ $$\Rightarrow \theta_1+\theta_2=\pi-2\phi$$

Now using sum-to-product:

$$\frac{\sin\theta_1+\sin\theta_2}{\cos\theta_1+\cos\theta_2} = \frac{2\sin\frac{\theta_1+\theta_2}{2}\cos\frac{\theta_1-\theta_2}{2}}{2\cos\frac{\theta_1+\theta_2}{2}\cos\frac{\theta_1-\theta_2}{2}} = \tan\frac{\theta_1+\theta_2}{2}$$ $$= \tan\left(\frac{\pi}{2}-\phi\right) = \cot\phi$$

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