Indian Navy SSR Trigonometry — practice questions

Square both sides of $\sin\theta + \cos\theta = 1$: $(\sin\theta + \cos\theta)^2 = 1^2$ $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1$ Using $\sin^2 + \cos^2 = 1$: $1 + 2\sin\theta\cos\theta = 1$ $\sin\theta \cos\theta = 0$. Verify: the original equation holds only when $\theta = 0°$ (so $\sin = 0$, $\cos = 1$) or $\theta = 90°$ (so $\sin = 1$, $\cos = 0$). In both cases the product is 0. ✓

Triple-angle for cosine: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. Derivation: $\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta = (2\cos^2\theta - 1)\cos\theta - 2\sin\theta\cos\theta \cdot \sin\theta = 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta = 2\cos^3\theta - \cos\theta - 2(1-\cos^2\theta)\cos\theta = 4\cos^3\theta - 3\cos\theta$. Companion: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ (option C — common trap). Mnemonic: in both, the coefficients are 3 and 4, and the cubed term has the same trig function as the angle outside.

Only sin⁻¹ and cos⁻¹ accept every value from −1 to 1. cosec⁻¹ and sec⁻¹ need |x| ≥ 1, so the open middle is cut out. tan⁻¹ and cot⁻¹ accept any real number. Source: syllabus Maths Part I, Ch 2 Inverse Trigonometric Functions.