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evaluate sqrt sin 4 x 4 cos 2 x sqrt cos 4 x 4 sin
Question:
Evaluate:
\[ \sqrt{\sin^4 x + 4\cos^2 x} - \sqrt{\cos^4 x + 4\sin^2 x} = ? \]
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Use trigonometric identities such as \( \cos 2x = 2\cos^2 x - 1 \) and square root simplifications to break complex expressions into standard forms.
TG EDCET - 2025
TG EDCET
Updated On:
Jun 9, 2025
\( \sin 2x \)
\( \cos 4x \)
\( \cos 2x \)
\( \sin 4x \)
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The Correct Option is
C
Solution and Explanation
We are asked to simplify the following expression: \[ \sqrt{\sin^4 x + 4\cos^2 x} - \sqrt{\cos^4 x + 4\sin^2 x} \] Let us first observe that: \[ \sin^4 x = (\sin^2 x)^2, \quad \cos^4 x = (\cos^2 x)^2 \] Start simplifying the first square root: \[ \sqrt{\sin^4 x + 4\cos^2 x} = \sqrt{(\sin^2 x)^2 + 4\cos^2 x} \] Now try substituting \( \sin^2 x = 1 - \cos^2 x \) and vice versa in the respective expressions: First term: \[ \sqrt{(1 - \cos^2 x)^2 + 4\cos^2 x} = \sqrt{1 - 2\cos^2 x + \cos^4 x + 4\cos^2 x} \] \[ = \sqrt{1 + 2\cos^2 x + \cos^4 x} \] Second term: \[ \sqrt{\cos^4 x + 4\sin^2 x} = \sqrt{(\cos^2 x)^2 + 4(1 - \cos^2 x)} = \sqrt{\cos^4 x + 4 - 4\cos^2 x} \] \[ = \sqrt{\cos^4 x - 4\cos^2 x + 4} \] Now simplify both: \[ \text{First term: } \sqrt{\cos^4 x + 2\cos^2 x + 1} = \sqrt{(\cos^2 x + 1)^2} = \cos^2 x + 1 \] \[ \text{Second term: } \sqrt{\cos^4 x - 4\cos^2 x + 4} = \sqrt{(\cos^2 x - 2)^2} = | \cos^2 x - 2 | \] Now, since \( 0 \leq \cos^2 x \leq 1 \Rightarrow \cos^2 x - 2 \leq 0 \Rightarrow | \cos^2 x - 2 | = 2 - \cos^2 x \) So: \[ \text{Expression} = (\cos^2 x + 1) - (2 - \cos^2 x) = \cos^2 x + 1 - 2 + \cos^2 x = 2\cos^2 x -1 \] Recall the identity: \[ \cos 2x = 2\cos^2 x - 1 \] Thus: \[ \sqrt{\sin^4 x + 4\cos^2 x} - \sqrt{\cos^4 x + 4\sin^2 x} = \cos 2x \]
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