1.1 Q-8

Question Statement

Prove the following identities:

i. $\quad \sinh 2x = 2 \sinh x \cdot \cosh x$

ii. $\quad \sec^2 x = 1 - \tanh^2 x$

iii. $\quad \operatorname{csch}^2 x = \coth^2 x - 1$

Background and Explanation

To solve these hyperbolic trigonometric identities, we need a basic understanding of the definitions and relationships of hyperbolic functions. Here are some key definitions:

• Hyperbolic sine: $\sinh x = \frac{e^x - e^{-x}}{2}$

• Hyperbolic cosine: $\cosh x = \frac{e^x + e^{-x}}{2}$

• Hyperbolic tangent: $\tanh x = \frac{\sinh x}{\cosh x}$

• Hyperbolic secant: $\operatorname{sech} x = \frac{1}{\cosh x}$

• Hyperbolic cosecant: $\operatorname{csch} x = \frac{1}{\sinh x}$

• Hyperbolic cotangent: $\operatorname{coth} x = \frac{\cosh x}{\sinh x}$

These formulas will help simplify and prove the identities.

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Solution

i. Prove that $\sinh 2x = 2 \sinh x \cdot \cosh x$

Start with the right-hand side (RHS):

$$\text{R.H.S} = 2 \sinh x \cdot \cosh x$$

Substitute the definitions of $\sinh x$ and $\cosh x$:

$$= 2 \left( \frac{e^x - e^{-x}}{2} \right) \left( \frac{e^x + e^{-x}}{2} \right)$$

Simplify:

$$= \frac{2}{4} \left( (e^x - e^{-x})(e^x + e^{-x}) \right)$$

Using the difference of squares formula:

$$= \frac{2}{4} \left( e^{2x} - e^{-2x} \right)$$

This simplifies to:

$$= \sinh 2x$$

Thus,

$$\text{L.H.S} = \sinh 2x = \text{R.H.S}$$

ii. Prove that $\sec^2 x = 1 - \tanh^2 x$

Start with the right-hand side (RHS):

$$\text{R.H.S} = 1 - \tanh^2 x$$

Substitute the definition of $\tanh x$:

$$= 1 - \left( \frac{e^x - e^{-x}}{e^x + e^{-x}} \right)^2$$

Simplify the expression:

$$= \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$

Now, simplify the numerator using the difference of squares formula:

$$= \frac{e^{2x} + 2 + e^{-2x} - (e^{2x} - 2 + e^{-2x})}{(e^x + e^{-x})^2}$$

Simplify the terms:

$$= \frac{4}{(e^x + e^{-x})^2}$$

Recognize this as the definition of $\operatorname{sech}^2 x$:

$$= \operatorname{sech}^2 x$$

Thus,

$$\text{L.H.S} = \operatorname{sech}^2 x = \text{R.H.S}$$

iii. Prove that $\operatorname{csch}^2 x = \coth^2 x - 1$

Start with the right-hand side (RHS):

$$\text{R.H.S} = \coth^2 x - 1$$

Substitute the definition of $\coth x$:

$$= \left( \frac{\cosh x}{\sinh x} \right)^2 - 1$$

Simplify:

$$= \frac{\cosh^2 x}{\sinh^2 x} - 1$$

Now, rewrite 1 as $\frac{\sinh^2 x}{\sinh^2 x}$:

$$= \frac{\cosh^2 x - \sinh^2 x}{\sinh^2 x}$$

Using the identity $\cosh^2 x - \sinh^2 x = 1$:

$$= \frac{1}{\sinh^2 x}$$

Recognize this as the definition of $\operatorname{csch}^2 x$:

$$= \operatorname{csch}^2 x$$

Thus,

$$\text{L.H.S} = \operatorname{csch}^2 x = \text{R.H.S}$$

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Key Formulas or Methods Used

• Hyperbolic functions: $\sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}, \quad \tanh x = \frac{\sinh x}{\cosh x}$

• Difference of squares formula: $(a - b)(a + b) = a^2 - b^2$

• Hyperbolic secant and cosecant: $\operatorname{sech} x = \frac{1}{\cosh x}, \quad \operatorname{csch} x = \frac{1}{\sinh x}$

• Identity: $\cosh^2 x - \sinh^2 x = 1$

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Summary of Steps

1. For i: Expand and simplify the expression for $2 \sinh x \cdot \cosh x$ to get $\sinh 2x$.
2. For ii: Expand $1 - \tanh^2 x$ and simplify to get $\operatorname{sech}^2 x$.
3. For iii: Simplify the expression for $\coth^2 x - 1$ to get $\operatorname{csch}^2 x$.