Trigonometric Ratio and Identities MCQ Questions & Answers in Trigonometry | Maths

The expression $$ = {\left( {\sin \frac{\pi }{{14}} \cdot \sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}} \right)^2} \cdot \sin \frac{{7\pi }}{{14}}.$$
Clearly, $$\sin\frac{\pi }{{14}} = \cos \left( {\frac{\pi }{2} - \frac{\pi }{{14}}} \right) = \cos \frac{{3\pi }}{7} = - \cos \left( {x - \frac{{3\pi }}{7}} \right) = - \cos \frac{{4\pi }}{7}.$$
$$\sin\frac{{3\pi }}{{14}} = \cos \frac{{2\pi }}{7},\sin \frac{{5\pi }}{7} = \cos \frac{\pi }{7}.$$
∴ value $$ = {\left( {\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7} \cdot \frac{{4\pi }}{7}} \right)^2} \cdot 1$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left\{ {\frac{{2\sin \frac{\pi }{7} \cdot \cos \frac{\pi }{7} \cdot \cos \frac{{2\pi }}{7} \cdot \cos \frac{{4\pi }}{7}}}{{2\sin \frac{\pi }{7}}}} \right\}^2} = ..... = \left( {\frac{{\sin \frac{{8\pi }}{7}}}{{8\sin \frac{\pi }{7}}}} \right) = \frac{1}{{{8^2}}}.$$

$$\eqalign{ & \because \,\,\cos \sqrt x {\text{ is non periodic}} \cr & \therefore \,\,{\text{cos}}\sqrt x + {\cos ^2}x\,{\text{can not be periodic}}{\text{.}} \cr} $$

Here, $$\cos x = \frac{{2\cos \left( {x - y} \right)\cos \left( {x + y} \right)}}{{\cos \left( {x - y} \right) + \cos \left( {x + y} \right)}} = \frac{{{{\cos }^2}x - {{\sin }^2}y}}{{\cos x \cdot \cos y}}$$
or, $${\cos ^2}x\left( {1 - \cos y} \right) = {\sin ^2}y\,\,\,\,{\text{or, }}{\cos ^2}x = 2{\cos ^2}\frac{y}{2}$$
or, $${\left( {\cos x \cdot \sec \frac{y}{2}} \right)^2} = 2.$$

$$\eqalign{ & {\sin ^3}x \cdot \sin 3x = \frac{{3\sin x - \sin 3x}}{4}\sin 3x \cr & {\sin ^3}x \cdot \sin 3x = \frac{3}{8}\left( {2\sin x \cdot \sin 3x} \right) - \frac{1}{4}{\sin ^2}3x \cr & {\sin ^3}x \cdot \sin 3x = \frac{3}{8}\left( {\cos 2x - \cos 4x} \right) - \frac{1}{8}\left( {1 - \cos 6x} \right) \cr & {\sin ^3}x \cdot \sin 3x = - \frac{1}{8} + \frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x. \cr} $$

$$\eqalign{ & {\cos ^6}x = 1 - {\sin ^4}x = {\cos ^2}x\left( {1 + {{\sin }^2}x} \right) \cr & \Rightarrow \,\,{\cos ^2}x = 0\,\,\,{\text{or, }}{\cos ^4}x = 1 + {\sin ^2}x \cr & 1 + {\sin ^2}x = {\cos ^4}x \cr & \Rightarrow \,\,{\sin ^2}x = - \left( {1 - {{\cos }^4}x} \right) = - {\sin ^2}x\left( {1 + {{\cos }^2}x} \right) \cr & \Rightarrow \,\,{\sin ^2}x = 0\,\,\,{\text{or, }}2 + {\cos ^2}x = 0\left( {{\text{absurd}}} \right). \cr & \therefore \,\,\cos x = 0\,\,\,{\text{or, }}\sin x = 0 \cr & \Rightarrow \,\,x = \frac{\pi }{2}, - \frac{\pi }{2},0,\pi , - \pi . \cr} $$

$$\eqalign{ & {\sec ^2}\theta + \sec \theta - \left( {\lambda + 1} \right) = 0 \cr & \therefore \,\,\sec \theta = \frac{{ - 1 \pm \sqrt {1 + 4\left( {\lambda + 1} \right)} }}{2} = \frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2}. \cr} $$
For real, $$\sec \theta ,4\lambda + 5 \geqslant 0,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \geqslant - \frac{5}{4}.$$
Also, $$\sec\theta \geqslant 1\,\,{\text{or, sec}}\theta \leqslant - {\text{1}}{\text{.}}$$
$$\eqalign{ & \therefore \,\,\frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2} \geqslant 1\,\,{\text{or, }}\frac{{ - 1 \pm \sqrt {4\lambda + 5} }}{2} \leqslant - 1 \cr & \Rightarrow \,\, - 1 + \sqrt {4\lambda + 5} \geqslant 2\,\,\,{\text{or, }} - 1 - \sqrt {4\lambda + 5} \leqslant - 2 \cr & \Rightarrow \,\,4\lambda + 5 \geqslant 9\,\,{\text{or, }}4\lambda + 5 \geqslant 1 \cr & \Rightarrow \,\,\lambda \geqslant 1\,\,\,{\text{or, }}\lambda \geqslant - 1 \cr & \therefore \,\,\lambda \geqslant - \frac{5}{4}\,{\text{and }}\lambda \geqslant 1\,\,\,{\text{or, }}\lambda \geqslant - \frac{5}{4}\,{\text{and }}\lambda \geqslant - 1 \cr & \therefore \,\,\lambda \geqslant 1\,\,{\text{or, }}\lambda \geqslant - 1 \cr & \therefore \,\,\lambda \in \left[ { - 1, + \infty } \right). \cr} $$

The given expression can be written as
$$\eqalign{ & \frac{{\left( {\cos 6x + \cos 4x} \right) + 5\left( {\cos 4x + \cos 2x} \right) + 10\left( {\cos 2x + 1} \right)}}{{\cos 5x + 5\cos 3x + 10\cos x}} \cr & = \frac{{2\cos 5x\cos x + 5 \cdot 2\cos 3x\cos x + 10 \cdot 2{{\cos }^2}x}}{{\cos 5x + 5\cos 3x + 10\cos x}} \cr & = \frac{{2\cos x\left( {\cos 5x + 5\cos 3x + 10\cos x} \right)}}{{\cos 5x + 5\cos 3x + 10\cos x}} = 2\cos x \cr} $$

We have
$$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - \frac{3}{2}$$
$$\eqalign{ & \Rightarrow \,\,2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] + 3 = 0 \cr & \Rightarrow \,\,2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] + {\sin ^2}\alpha + {\cos ^2}\alpha + {\sin ^2}\beta + {\cos ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\gamma = 0 \cr & \Rightarrow \,\,\left[ {{{\sin }^2}\alpha + {{\sin }^2}\beta + {{\sin }^2}\gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha } \right] + \left[ {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha } \right] = 0 \cr & \Rightarrow \,\,{\left[ {\sin \alpha + \sin \beta + \sin \gamma } \right]^2} + {\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = 0 \cr & \Rightarrow \,\,\sin \alpha + \sin \beta + \sin \gamma = 0\,\,{\text{and }}\cos \alpha + \cos \beta + \cos \gamma = 0 \cr} $$
∴ A and B both are true.

$$\eqalign{ & \sum\limits_{k = 1}^{13} {\frac{1}{{\sin \left( {\frac{\pi }{4} + \frac{{\left( {k - 1} \right)\pi }}{6}} \right)\sin \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)}}} \cr & = \sum\limits_{k = 1}^{13} {\frac{1}{{\sin \frac{\pi }{6}}}\left[ {\frac{{\sin \left\{ {\frac{\pi }{4} + \frac{{k\pi }}{6} - \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right)} \right\}}}{{\sin \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right)\sin \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)}}} \right]} \sum\limits_{k = 1}^{13} {2\left[ {\cot \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)} \right]} \cr & = 2\left[ {\left\{ {\cot \frac{\pi }{4} - \cot \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right)} \right\} + \left\{ {\cot \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{2\pi }}{6}} \right)} \right\} + ..... + \left\{ {\cot \left( {\frac{\pi }{4} + \frac{{12\pi }}{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{13\pi }}{6}} \right)} \right\}} \right] \cr & = 2\left[ {\cot \frac{\pi }{4} - \cot \left( {\frac{\pi }{4} + \frac{{13\pi }}{6}} \right)} \right] \cr & = 2\left[ {1 - \cot \frac{{5\pi }}{{12}}} \right] \cr & = 2\left[ {1 - \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right] \cr & = 2\left[ {1 - \left( {2 - \sqrt 3 } \right)} \right] \cr & = 2\left( {\sqrt 3 - 1} \right). \cr} $$

As $${\cos ^2}x + {\sin ^2}x = 1,{\sin^2}x \geqslant {\sin ^4}x$$       (equality holding when $$\sin x = 0$$  )
and $${\cos^2}x \geqslant {\cos ^7}x$$   (equality holding when $$\cos x = 0$$  )
we find that $${\cos ^7}x + {\sin ^4}x = 1$$    can hold only when
$$\cos x = 0\,\,\,\,{\text{or, }}\sin x = 0\left( {x \ne \pi , - \pi } \right).$$
So, $$x = - \frac{\pi }{2},\frac{\pi }{2},0.$$