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Find out the range of function 4\sin x -\sin^2 x -1

Option: 1

[-1,2]


Option: 2

[-1,4]


Option: 3

[-6,2]


Option: 4

[-6,1]


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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\\4\sin x -\sin^2 x -1=-(\sin x-2)^2+3\\ \text{Range of this function is [-6,2]}

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Rishi

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