Inverse circular functions,Principal values of ${s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x$ .
${t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {t a n}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
(a) ${c o s}^{- 1} (c o s \frac{7 π}{6})$
(b) ${c o s}^{- 1} (c o s \frac{4 π}{3})$

Open in App

Solution

(a) ${c o s}^{- 1} c o s (\frac{7 π}{6}) = \frac{7 π}{6}$ is wrong, as $7 π / 6$ is greater than $π$ .
Here $\frac{7 π}{6} = 2 π - \frac{5 π}{6}$ so that $\frac{5 π}{6}$ lies between $0$ and $π$
$∴$ ${c o s}^{- 1} c o s (\frac{7 π}{6}) = {c o s}^{- 1} c o s (2 π - \frac{5 π}{6})$
$= {c o s}^{- 1} c o s (\frac{5 π}{6}) = \frac{5 π}{6}$ .
(b) ${c o s}^{- 1} c o s \frac{4 π}{3} = {c o s}^{- 1} c o s (2 π - \frac{2 π}{3})$
$= {c o s}^{- 1} c o s \frac{2 π}{3} = \frac{2 π}{3}$ .

Suggest Corrections

Similar questions

Q. Inverse circular functions,Principal values of

{sin}^{- 1} x, {cos}^{- 1} x, {tan}^{- 1} x

{tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
Prove that

{tan}^{- 1} \frac{1 - x}{1 + x} {tan}^{- 1} \frac{1 - y}{1 + y} = {sin}^{- 1} \frac{y - x}{\sqrt{1 + x^{2}} \sqrt{1 + y^{2}}}

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

2 {t a n}^{- 1} (\sqrt{\frac{a - b}{a + b}} t a n \frac{θ}{2}) = {c o s}^{- 1} \frac{a c o s θ + b}{a + b c o s θ}

Inverse circular functions,Principal values of ${sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x$ .
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
Evaluate the following :
(a) $sin [\frac{π}{3} - {sin}^{- 1} (- \frac{1}{2})]$
(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$