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Mathematics

If sin 3A = 1 and 0 ≤ A ≤ 90°, find :

(i) sin A

(ii) cos 2 A

(iii) tan2 A - 1cos2 A\dfrac{1}{\text{cos}^2 \text{ A}}

Trigonometric Identities

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Answer

sin 3A = 1

⇒ sin 3A = sin 90°

So, 3A = 90°

⇒ A = 90°3=30°\dfrac{90°}{3} = 30°

(i) sin A = sin 30° = 12\dfrac{1}{2}

Hence, sin A = 12\dfrac{1}{2}.

(ii) cos 2 A

= cos (2 x 30°)

= cos 60°

= 12\dfrac{1}{2}

Hence, cos 2A = 12\dfrac{1}{2}.

(iii) tan2 A - 1cos2 A\dfrac{1}{\text{cos}^2 \text{ A}}

=tan230°−1cos230°=(13)2−1(32)2=13−134=13−43=1−43=−33=−1= \text{tan}^2 \text{30°} - \dfrac{1}{\text{cos}^2 \text{30°}}\\[1em] = \Big(\dfrac{1}{\sqrt3}\Big)^2 - \dfrac{1}{\Big(\dfrac{\sqrt3}{2}\Big)^2}\\[1em] = \dfrac{1}{3} - \dfrac{1}{\dfrac{3}{4}}\\[1em] = \dfrac{1}{3} - \dfrac{4}{3}\\[1em] = \dfrac{1 - 4}{3}\\[1em] = \dfrac{- 3}{3}\\[1em] = - 1

Hence, tan2 A - 1cos2 A\dfrac{1}{\text{cos}^2 \text{ A}} = -1.

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