Question

find the minimum value of sin 2x-root3 cos 2x

Answer 1

Answer:

HOPE THIS BELOW ONE HELPS YOU........

Step-by-step explanation:

Let say A = sin 2x - cos 2x

Squaring both sides

=> A² = ( sin 2x - cos 2x )²

=> A² = Sin²2x + Cos²2x - 2Sin2xCos2x

=> A² = 1 - Sin4x

-1 ≤ Sin4x ≤ 1

=> A² = 1 - (-1 ) = A = ± √2

Sin4x = -1

=> A² = 0 => A = 0

Min Value = -√2 & Max Value = √2

Answer 2

Minimum value of sin2x - √3 cos 2x is -2

Given problem:

Find the minimum value of sin2x - √3 cos 2x

To find:

Minimum value of sin2x - √3 cos 2x

Solution:

The minimum value of a sinx + b cosx is is given by $- \sqrt{a^{2}+b^{2} }$  

Compare given function sin2x - √3 cos 2x with a sinx + b cosx

⇒ a = 1 and  b = -√3

⇒ $- \sqrt{a^{2}+b^{2} } = - (\sqrt{1^{2} + (\sqrt{3})^{2} }$

⇒ $- (\sqrt{1 + 3 }) = - \sqrt{4} = - 2$

Therefore, Minimum value of sin2x - √3 cos 2x is -2

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