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Find the range of function f(x)=1/(2sin(X)-3)

Option: 1

[-1,-\frac{1}{5}]


Option: 2

[-1,0]


Option: 3

[-1,-\frac{1}{4}]


Option: 4

[-1,-\frac{1}{6}]


Answers (1)

best_answer

 

Sign of Trigonometric Function -

Sign of Trigonometric Function-

 

 

The sign of trigonometric ratios of an angle depends on the quadrant in which the terminal side of the angles lies. We always take OP = r as positive. Thus, the sign of trigonometric functions depends on the sign of x and/or y.  

 

 

An angle is said to be in that quadrant in which it's terminal ray lies (here terminal ray is OP).

  1. In the first quadrant x and y are positive so sin θ, cos θ, tan θ, sec θ, csc θ, and cot θ are all positive.

  2. In the second quadrant, x is negative and y is positive so only sin θ and cosec θ are positive.

  3. In the third quadrant, x is negative and y is negative so only tan θ and cot θ are positive.

  4. In the fourth quadrant, x is positive and y is negative so only cos θ and sec θ are positive.

 

 

To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class” or “All Sugar To Coffee.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise.

Depending on the sign of x and y, the various trigonometric ratio will have different sign given.

 

  • sin (-θ) = - sin (θ)

  • cos (-θ) = cos (θ)

  • csc (-θ) = - csc (θ)

  • sec (-θ) = sec (θ)

  • tan (-θ) = - tan (θ)

  • cot (-θ) = -cot (θ)


\begin{array}{l}{\sin (-\theta)=\frac{-y}{r}=-\frac{y}{r}=-\sin \theta} \\\ {\cos (-\theta)=\frac{x}{r}=\cos \theta, }\\\tan (-\theta)=-\frac{y}{x}=\tan \theta \\ {\text { Taking the reciprocals of these\;trigonometric ratios, we have }} \\ {\csc (-\theta)} {=-\csc \theta, \;\;\;\;\;\;\;\sec (-\theta)=\sec \theta \text { \;\;\;\;and } } \\ {\cot (-\theta)=-\cot \theta }\end{array}

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\begin{array}{l}{\text f(x)=\frac{1}{2 \sin x-{3}}} \\ {\text { -1 } \leq \sin x \leq 1} \\ {\Rightarrow-2 \leq 2 \sin x \leq 2} \\ {\Rightarrow-2-{3} \leq 2 \sin x-{3} \leq 2-{3}} \\ \Rightarrow \frac{1)}{f(x)} \leq -{1} \\ \Rightarrow f(x) \epsilon [-1,0)\\ \Rightarrow \frac{1)}{f(x)} \geq -5 \\ \Rightarrow f(x)\epsilon (-\infty,\frac{-1}{5}]\ U(0,\infty) by\ both\ inequalities\ f(x)\epsilon [-1,-\frac{1}{5}]\end{array}

Posted by

Ritika Harsh

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