Domain and range of trigonometric functions.

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Solution

sin(x) • Domain: R = (−∞,∞)
•Range: [−1, 1]
•Period: 2π
cos(x) • Domain: R
• Range: [−1, 1]
• Period: 2π
tan(x) • Domain: {x|x 6= π 2 + kπ, k = ..., −1, 0, 1, ...} = {x|x 6= ..., − 3π 2 , − π 2 , π 2 , 3π 2 , ...}
• Range: R
• Period: π

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\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

s i n^{- 1} x = | x - a |

\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0

How to find the range and domain of any trigonometric ratio

(sin, cos, tan)

f (x) = {sin}^{20} x + {cos}^{48} x

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