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find the minimum value of \sin 2x+\csc 2x \ \ x\epsilon [0,\frac{\pi}{2}]

Option: 1

-\infty


Option: 2

-2


Option: 3

2


Option: 4

1


Answers (1)

best_answer

Maximum and Minimum value of Trigonometric Function -

Maximum and Minimum value of Trigonometric Function

 

We know that range of sin x and cos x which is [-1, 1],  

If there is a trigonometric function in the form of a sin x + b cos x, then replace a with r cos ? and b with r sin ?.

Then we have,\\\mathrm{\mathit{a}\sin x+\mathit{b}\cos x =\mathit{r}\cos\theta \sin x+\mathit{r}\sin \theta \cos x}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\mathit{r}\left ( \cos\theta \sin x+\sin \theta \cos x \right )}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\mathit{r} \sin(x+\theta)}\\\mathrm{where\;\;\mathit{r}=\sqrt{\mathit{a^2+b^2}}\;\;and,\;\;\tan\theta=\frac{b}{a}}

 

\\\mathrm{Since,\;\;-1\leq\sin(x+\theta)\leq1}\\\mathrm{Multiply\;with\;'r'}\\\mathrm{\Rightarrow -r\leq\sin(x+\theta)\leq r}\\\mathrm{\Rightarrow -\sqrt{\mathit{a^2+b^2}}\leq\sin(x+\theta)\leq\sqrt{\mathit{a^2+b^2}}}\\\mathrm{\Rightarrow -\sqrt{\mathit{a^2+b^2}}\leq \mathit{a}\sin x+\mathit{b}\cos x\leq\sqrt{\mathit{a^2+b^2}}}

So, the minimum value of trigonometric function a sin x + b cos x is -\sqrt{\mathit{a^2+b^2}} and maximum value is\sqrt{\mathit{a^2+b^2}}.

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\text{ we know } A.M. \geq G.M. \\ \frac{\sin 2x+\csc 2x}{2}\geq (\sin 2x \cdot\csc 2x)^\frac{1}{2}\\ {\sin 2x+\csc 2x}\geq 2

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himanshu.meshram

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