Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Hyperbolic Functions Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Hyperbolic Functions Important Questions

Question 1.

Prove that for any x ∈ R, sinh (3x) = 3 sinh x + 4sinh x

Answer:

LHS = sinh (3x)

= sinh (2x + x)

= sinh (2x) . cosh(x) + cosh (2x) . sinh (x) = (2sinh x cosh x)cosh x (1 +2sinh^{2 }x)sinh x

= 2 sinh x (cosh^{2} x) + (1 + 2 sinh^{2} x) sinh x

= 2 sinh x (1 + sinh^{2} x) + (1 + 2 sinh^{2} x) sinh x

∵ cosh^{2} x – sinh^{2} x = 1

= 3 sinh x + 4 sinh^{3} x

∴ sinh (3x) = 3 sinh x + 4 sinh^{3} x

Question 2.

Prove that for any x ∈ R, tanh 3x = \(\frac{3 \tanh x+\tanh ^{3} x}{1+3 \tanh ^{2} x}\)

Answer:

tanh 3x = tan (2x + x)

Question 3.

If cosh x = \(\frac{5}{2}\), find the values of

i) cosh (2x) and

ii) sinh (2x)

Answer:

cosh (x) = \(\frac{5}{2}\)

(i) cosh (2x) = 2 cosh^{2} (x) – 1

= 2(\(\frac{5}{2}\))^{2} – 1 = \(\frac{25}{2}\) – 1 = \(\frac{23}{2}\)

ii) sinh^{2} (2x) = cosh^{2} (2x) – 1

= (\(\frac{23}{2}\))^{2} – 1 = \(\frac{529-4}{2}\) = \(\frac{525}{4}\)

∴ sinh (2x) = ±\(\sqrt{\frac{525}{4}}\) = ±\(\frac{5 \sqrt{21}}{2}\)

Question 4.

If coshx = sec θ then prove that tan h^{2}\(\frac{x}{2}\) = tan^{2}\(\frac{\theta}{2}\)

Answer:

tan h^{2}\(\frac{x}{2}\) = \(\frac{\cosh x-1}{\cosh x+1}\)

= \(\frac{\sec \theta-1}{\sec \theta+1}\) = \(\frac{1-\cos \theta}{1+\cos \theta}\) = tan^{2}\(\frac{\theta}{2}\)

Question 5.

If θ ∈ (-\(\frac{\pi}{4}\), \(\frac{\pi}{4}\)) and x = log_{e}(cot(\(\frac{\pi}{4}\) + θ) then prove that

i) cosh x = sec 2θ and

ii) sinh x = -tan 2θ

Answer:

Question 6.

If sinh x = 5, show that x = log_{e} (5 + \(\sqrt{26}\))

Answer:

∴ sinh (x) = 5

⇒ x = sinh^{-1} (5)

= log_{e} (5 + \(\sqrt{5^{2}+1}\))

= log_{e} (5 + \(\sqrt{26}\))

[sin^{-1} (x) = log_{e} (x + \(\sqrt{x^{2}+1}\)) for all x ∈ R]

Question 7.

Show that tanh^{-1}(\(\frac{1}{2}\)) = \(\frac{1}{2}\) log_{e}3 (A.P) [Mar 15; May 07, 05; Mar 08, 05]

Answer:

∵tanh^{-1} (x) = \(\frac{1}{2}\)log_{e}(\(\frac{1+x}{1-x}\)) for all x ∈ (-1, 1)

∵ tanh^{-1} (\(\frac{1}{2}\)) = \(\frac{1}{2}\)log_{e}(\(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\))

= \(\frac{1}{2}\)log_{e}(3)