Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If 12 Sin X − 9sin2 X Attains Its Maximum Value at X = α, Then Write the Value of Sin α. - Mathematics

Short Note

If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.

#### Solution

$\text{ Let } f\left( x \right) = 12\sin x - 9 \sin^2 x$

$= - \left( 9 \sin^2 x - 12 \sin x \right)$

$= - \left[ \left( 3\sin x \right)^2 - 2 . 3 \sin x . 2 + 2^2 - 4 \right]$

$= - \left[ \left( 3 \sin x - 2 \right)^2 - 4 \right]$

$= 4 - \left( 3 \sin x - 2 \right)^2$

$\text{ Minimum value of } \left( 3 \sin x - 2 \right)^2 \text{ is } 0 .$

$\text{ Therefore, maximum value of f }\left( x \right) = 4 - \left( 3 \sin x - 2 \right)^2 \text{ is } 4 .$

$\text{ We are given that } 12\sin x - 9 \sin^2 x \text{ will attain its maximum value at } x = \alpha .$

$\therefore 12\sin\alpha - 9 \sin^2 \alpha = 4$

$\Rightarrow - 9 \sin^2 \alpha + 12\sin\alpha - 4 = 0$

$\Rightarrow 9 \sin^2 \alpha - 12 \sin\alpha + 4 = 0$

$\Rightarrow 9 \sin^2 \alpha - 6\sin\alpha - 6\sin\alpha + 4 = 0$

$\Rightarrow 3\sin\alpha\left( 3\sin\alpha - 2 \right) - 2\left( 3\sin\alpha - 2 \right) = 0$

$\Rightarrow \left( 3\sin\alpha - 2 \right)\left( 3\sin\alpha - 2 \right) = 0$

$\therefore \sin\alpha = \frac{2}{3}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Q 5 | Page 26