Question 10, Exercise 8.1

Solutions of Question 10 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Verify: $\sin \left(\dfrac{\pi}{2}-\alpha\right)=\cos \alpha$

Solution.

$$\begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha\right) \\ & =\sin\frac{\pi}{2} \cos \alpha - \cos \frac{\pi}{2} \sin\alpha \\ & = 1\times \cos \alpha - 0 \times \sin\alpha \\ & = \cos\alpha = R.H.S \end{align*}$$

Verify: $\cos (\pi-\alpha)=-\cos \alpha$

Solution.

$$\begin{align*} L.H.S & = \cos(\pi - \alpha) \\ & = \cos \pi \cos \alpha + \sin \pi \sin \alpha \\ & = (-1) \cdot \cos \alpha + 0 \cdot \sin \alpha \\ & = -\cos \alpha \\ & = R.H.S. \end{align*}$$

Verify: $\cos \left(\alpha+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\cos \alpha-\sin \alpha)$

Solution.

$$\begin{align*} L.H.S &= \cos \left(\alpha+\frac{\pi}{4}\right) \\ & = \cos\alpha \cos\frac{\pi}{4} - \sin\alpha \sin\frac{\pi}{4} \\ & = \cos\alpha\cdot \frac{1}{\sqrt{2}} - \sin\alpha\cdot \frac{1}{\sqrt{2}}\\ & = \frac{1}{\sqrt{2}}(\cos \alpha-\sin \alpha)\\ & = R.H.S \end{align*}$$

Verify: $\sin \left(\beta+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}(\cos \beta+\sin \beta)$

Solution.

$$\begin{align*} L.H.S & = \cos \left(\alpha + \frac{\pi}{4}\right) \\ & = \cos \alpha \cos \frac{\pi}{4} - \sin \alpha \sin \frac{\pi}{4} \\ & = \cos \alpha \cdot \frac{1}{\sqrt{2}} - \sin \alpha \cdot \frac{1}{\sqrt{2}} \\ & = \frac{1}{\sqrt{2}} \left( \cos \alpha - \sin \alpha \right) \\ & = R.H.S. \end{align*}$$

Verify: $\tan \left(\gamma-\frac{\pi}{4}\right)=\frac{\tan \gamma-1}{\tan \gamma+1}$

Solution.

$$\begin{align*} L.H.S & = \tan \left(\gamma - \frac{\pi}{4}\right) \\ & = \frac{\tan \gamma - \tan \frac{\pi}{4}}{1 + \tan \gamma \tan \frac{\pi}{4}} \\ & = \frac{\tan \gamma - 1}{1 + \tan \gamma \cdot 1} \quad (\text{since } \tan \frac{\pi}{4} = 1) \\ & = \frac{\tan \gamma - 1}{1 + \tan \gamma} \\ & = R.H.S. \end{align*}$$

Verify: $\tan \left(\gamma+\frac{\pi}{4}\right)=\frac{1+\tan \gamma}{1-\tan \gamma}=\frac{\cos \gamma+\sin \gamma}{\cos \gamma-\sin \gamma}$

Solution.

$$\begin{align*} L.H.S & = \tan \left(\gamma + \frac{\pi}{4}\right) \\ & = \frac{\tan \gamma + \tan \frac{\pi}{4}}{1 - \tan \gamma \tan \frac{\pi}{4}} \\ & = \frac{\tan \gamma + 1}{1 - \tan \gamma \cdot 1} \quad (\text{since } \tan \frac{\pi}{4} = 1) \\ & = \frac{\tan \gamma + 1}{1 - \tan \gamma} ..... (1) \\ & = \frac{\sin \gamma / \cos \gamma + 1}{1 - \sin \gamma / \cos \gamma} \\ & = \frac{\frac{\sin \gamma + \cos \gamma}{\cos \gamma}}{\frac{\cos \gamma - \sin \gamma}{\cos \gamma}} \\ & = \frac{\cos \gamma + \sin \gamma}{\cos \gamma - \sin \gamma}..... (2) \end{align*}$$ Combining L.H.S with (1) and (2), we have $$\tan \left(\gamma+\frac{\pi}{4}\right)=\frac{1+\tan \gamma}{1-\tan \gamma}=\frac{\cos \gamma+\sin \gamma}{\cos \gamma-\sin \gamma}.$$

Verify: $\cos (x+y)+\cos (x-y)=2 \cos x \cos y$

Solution.

$$\begin{align*} L.H.S & = \cos (x+y) + \cos (x-y) \\ & = \cos x \cos y - \sin x \sin y + \cos x \cos y + \sin x \sin y \\ & = 2 \cos x \cos y \\ & = R.H.S. \end{align*}$$

Verify: $\sin (x+y)-\sin (x-y)=2 \cos x \sin y$

Solution.

$$\begin{align*} L.H.S & = \sin (x+y) - \sin (x-y) \\ & = \left(\sin x \cos y + \cos x \sin y\right) - \left(\sin x \cos y - \cos x \sin y\right) \\ & = \sin x \cos y + \cos x \sin y - \sin x \cos y + \cos x \sin y \\ & = 2 \cos x \sin y \\ & = R.H.S. \end{align*}$$

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