Question

$\huge\sf{Question}$
If 4 cos² x0 – 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin2 x0 + cos2 x0

(iii) cos2 x0 – sin2 x0.

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Answer 1

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Answer 2

$\huge\color{pink}{Answer}$

If 4 cos² x° – 1 = 0 and 0 ≤ x ≤ 90, then :-

(i) x = 60°

$Explanation: - \\ 4cos²x° = 0 \\ cos²x° = \frac{1}{4} \\ cos²x° = (\frac{1}{2} {)}^{2} \\ cosx° = cos60° \\ x = 60°$

(ii) sin²x° + cos²x° = 1

$Explanation : - \\ The \: value \: of \: x = 60° so, \\ = sin² 60° + cos²60° \\ = (\frac{ \sqrt{3} }{2} {)}^{2} + (\frac{1}{2} {)}^{2} \\ = \frac{3}{4} + \frac{1}{4} \\ \frac{4}{4} = 1$

(iii) cos²x° - sin²x° = -1/2

$Explanation: - \\ Since \: the \: value \: of \: x = 60° so, \\ =cos²60° - sin²60° \\ (\frac{1}{2} {)}^{2} - ( \frac{ \sqrt{3} }{2} {)}^{2} \\ \frac{1}{4} - ( \frac{3}{4} ) \\ \frac{1}{4} - \frac{3}{4} \\ \frac{ - 2}{4} = \frac{ - 1}{2}$

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