If 4 cos2 x° – 1 = 0 and 0 ≤ x ≤ 90, find

If 4 cos² x° – 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin² x° + cos² x°

(iii) cos² x° – sin² x°

Answer :

4 cos² x° – 1 = 0

⇒ 4cos² x° = 1

cos² x° = ¼

⇒ cos x° = ± √1/4

⇒ cos x° = + √1/4 [0 ≤ x ≤ 90°, then cos x° is positive]

⇒ cos x° = ½

cos 60° = ½

cos x° = cos 60°

x = 60

(ii) sin² x° + cos² x° = sin² 60° + cos² 60°

= (√3/2)² + (1/2)²

= ¾ + ¼

= (3 + 1)/4

= 4/4

= 1

Hence, sin² x° + cos² x° = 1.

(iii) cos² x° – sin² x° = cos² 60° – sin² 60°

= (1/2)² – (√3/2)²

= ¼ – (√3/2 × √3/2)

= ¼ – ¾

= (1 – 3)/4

= -2/4

= – ½

Hence, cos² x° – sin² x° = – ½.

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