Use these Inter 1st Year Maths 1A Formulas PDF Chapter 9 Hyperbolic Functions to solve questions creatively.

## Intermediate 1st Year Maths 1A Hyperbolic Functions Formulas

→ ex = 1 + $$\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots+\frac{x^{n}}{n !}$$+ … ∞

→ sinhx = $$\frac{e^{x}-e^{-x}}{2}$$
and cosh x = $$\frac{e^{x}+e^{-x}}{2}$$

→ tanh x = $$\frac{\sinh x}{\cosh x}$$,

→ coth x = $$\frac{1}{\tanh x}$$,

→ sech x = $$\frac{1}{\cosh x}$$

→ cosech x = $$\frac{1}{\sinh x}$$, if x ≠ 0

→ cosh2x – sinh2x – 1, sech2x – 1 – tanh2x, cosech2x = coth2x – 1

 Function y = f(x) Domain (x) Range (y) sinh x R R cosh x R (1, ∞) tanh x R (1, 1) coth x R- {0} (-∞, -1) ∪ (1, ∞) sech x R (0, 1] cosech x R – {0} R – {0}

→ sinh-1x = loge [x + $$\sqrt{x^{2}+1}$$ ] for all x ∈ R

→ cosh-1x = loge [x + $$\sqrt{x^{2}-1}$$] for all x ∈ (1, ∞)

→ tanh-1x = $$\frac{1}{2}$$loge$$\left(\frac{1+x}{1-x}\right)$$, |x| < 1 (i.e) for all x ∈ (-1, -1)

→ coth-1x = $$\frac{1}{2}$$loge$$\left(\frac{1+x}{1-x}\right)$$, |x| > 1 (i.e) for all x ∈ (-∞, -1) (1, ∞)

→ sech-1x = loge$$\left(\frac{1+\sqrt{1+x^{2}}}{x}\right)$$ for all x ∈ (0, 1)

→ cosech-1x = loge$$\left(\frac{1-\sqrt{1+x^{2}}}{x}\right)$$, if x < 0 (i.e,) x ∈ (-∞, 0) and
= loge$$\left(\frac{1+\sqrt{1+x^{2}}}{x}\right)$$, if x > 0 (i.e,) x ∈ (0, ∞)

→ sinh (x + y) = sinh x cosh y + cosh x sinh y

→ cosh (x + y)= cosh x cosh y + sinh x sinh y

→ sinh (x – y) = sinh x cosh y – cosh x sinh y

→ cosh (x – y) = cosh x cosh y – sinh x sinh y

→ tanh (x + y) = $$\frac{\tanh x+\tanh y}{1+\tanh x \tanh y}$$

→ tanh (x – y) = $$\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}$$

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

→ cosh 2x = cosh2x + sinh2x
= 2 cosh2x – 1 = 1 + 2 sinh2 x = $$\frac{1+\tanh ^{2} x}{1-\tanh ^{2} x}$$

→ tanh 2x = $$\frac{2 \tanh x}{1+\tanh ^{2} x}$$

→ sinh 3x = 3sinh x + 4sinh3 x

→ cosh 3x = 4cosh3 x – 3cosh x

→ tanh 3x = \frac{3 \tanh x+\tanh ^{3} x}{1+3 \tanh ^{2} x}

→ Inverse hyperbolic functions:

 Function y = f(x) Domain (x) Range (y) (i) sinh-1(x) IR IR (ii) cosh-1(x) [1, ∞) [0, ∞) (iii) tanh-1(x) (-1, 1) IR (iv) coth-1(x) R –[-1, 1] R- {0} (v) sech-1(x) (0, 1] [0, ∞) (vi) cosech-1(x) R- {0} R – {0}

→ sin hx = $$\frac{e^{x}-e^{-x}}{2}$$

→ cos hx = $$\frac{e^{x}+e^{-x}}{2}$$

→ tan hx = $$\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

→ cosec hx = $$\frac{2}{e^{x}-e^{-x}}$$

→ sec hx = $$\frac{2}{e^{x}+e^{-x}}$$

→ cot hx = $$\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$$

→ cos h2x – sinh2x = 1

→ 1 – tanh2x = sech2x

→ cot h2x – 1 = cosech2x

→ Prove that sinh-1x = log{x + $$\sqrt{x^{2}+1}$$}
Proof:
Let sinh-1x = y ⇒ x = sinh y

→ Prove that cosh-1x = loge(x – $$\sqrt{x^{2}-1}$$
proof:
Let cosh-1x = y ⇒ x = cosh y

→ Prove that Tan-1x = $$\frac{1}{2}$$loge$$\left(\frac{4 x}{1-x}\right)$$
proof:
Let tanh-1x = y ⇒ x = tanh y

→ sech-1x = log$$\left\{\frac{1+\sqrt{1-x^{2}}}{x}\right\}$$

→ cosech-1x = log$$\left(\frac{1-\sqrt{1+x^{2}}}{x}\right)$$ x < 0 = log$$\left\{\frac{1-\sqrt{1+x^{2}}}{x}\right\}$$ x > 0

→ sin h(x + y) = sin hx cos hy + cos hx sin hy

→ sinh(x – y) = sin hx cos hy + cos hx sin hy

→ cosh(x + y) = cos hx cos hy + sin hx sin hv

→ cosh(x – y) = cos hx cos hy – sin hx sin hy

→ sin h2x = 2 sin hx cos hx = 

→ cos h2x = cosh2x + sinh2x = 2cosh2x – 1 = 1 + 2sinh2x = $$\frac{1+{Tanh}^{2} x}{1-{Tanh}^{2} x}$$

→ Tanh(x + y) = $$\frac{\text { Tanhx+Tanhy }}{1-\text { TanhxTanhy }}$$

→ Tanh(x – y) = $$\frac{\text { Tanhx-Tanhy }}{1+\text { TanhxTanhy }}$$

→ coth(x + y) = $$\frac{\cot h x \cot h y+1}{\cot h y+\cot h x}$$

→ coth(x – y) = $$\frac{\cot h x \cot h y-1}{\cot h y-\cot h x}$$

→ Tanh2x = $$\frac{2 \tan h x}{1+\tanh ^{2} x}$$

→ cot h2x = $$\frac{\operatorname{coth}^{2} x+1}{2 \cot h x}$$