Determine the principal value of cos-1-1-2-

Determining the principal value of cos^ 1( 1/2)Let, y=cos^ 1( 1/2)cos(y)=( 1/2)cos(y)= cos(π/3)cos(y)=cos(π π/3)cos(y)=cos(2π/3)Now we know that,Range of the pr ...

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Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Determine the...

Question

Determine the principal value of $\cos^{- 1} (- \frac{1}{2})$ .

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Solution

Determining the principal value of $\cos^{- 1} (- \frac{1}{2})$
Let, $y = \cos^{- 1} (- \frac{1}{2})$
$\cos (y) = (- \frac{1}{2})$
$\cos (y) = - \cos (\frac{π}{3})$
$\cos (y) = \cos (π - \frac{π}{3})$
$\cos (y) = \cos (\frac{2 π}{3})$
Now we know that,
Range of the principal value of $\cos^{- 1} (x) = [0, π]$
Hence, the principal value of $\cos^{- 1} (- \frac{1}{2}) = (\frac{2 π}{3})$ .

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Similar questions

Q. If

f (x) = {sin}^{- 1} {\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}}, - \frac{1}{2} \leq x \leq 1,

then

f (x)

is equal to

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. If

f (x) = {sin}^{- 1} {\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}}

- \frac{1}{2} \leq x \leq 1

, then

f (x)

is equal to :

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

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Determine the principal value of cos-1-12.

Determining the principal value of cos-1-12Let, y=cos-1-12cosy=-12cosy=-cosπ3cosy=cosπ-π3cosy=cos2π3Now we know that,Range of the principal value of cos-1x=[0,π]Hence, the principal value of cos-1-12=2π3.

Determine the principal value of $\cos^{- 1} (- \frac{1}{2})$ .