Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

Answer: Option C

Let x = 2sin²θ + 3cos²θ
⇒ x = 2sin²θ + 2cos²θ + cos²θ
⇒ x = 2(sin²θ + cos²θ) + cos²θ
⇒ x = 2 + cos²θ     [since sin²θ + cos²θ = 1]
Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2

Alternative Solution:
2sin²θ + 3cos²θ
Minimum value is 2,
[If x sin²θ + y cos²θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]

Answer: Option B

$$\eqalign{ & \left( {2\sin \theta + 3\cos \theta } \right) \cr & {\text{Maximum value of}} \cr & a\sin \theta + b\cos \theta \cr & = \sqrt {{a^2} + {b^2}} \cr & = \sqrt {{2^2} + {3^2}} \cr & = \sqrt {4 + 9} \cr & = \sqrt {13} \cr} $$

Answer: Option C

$$\eqalign{ & {\cos ^2}\theta = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & {\text{Max value of }}{\cos ^2}\theta = 1 \cr & \Rightarrow 1 = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & \Rightarrow 4xy = {\left( {x + y} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow 0 = {x^2} + {y^2} - 2xy \cr & \Rightarrow 0 = {\left( {x - y} \right)^2} \cr & \Rightarrow 0 = x - y \cr & \Rightarrow x = y \cr} $$

Answer: Option D

$$\eqalign{ & {\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & {\text{Squaring both sides}} \cr & {\sin ^4}\theta + {\cos ^4}\theta \cr & = 1 - 2{\sin ^2}\theta . {\cos ^2}\theta \cr & {\text{Put}}\,\theta = {90^ \circ } \cr & = 1 - 2{\sin ^2}{90^ \circ } \times {\cos ^2}{90^ \circ } \cr & = 1 - 0 \cr & = 1 \cr} $$

Answer: Option C

$$\eqalign{ & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow \cos \theta > {\cos ^2}\theta \cr & \Rightarrow \cos {60^ \circ } > {\cos ^2}{60^ \circ } \cr & \Rightarrow \frac{1}{2} > \frac{1}{4} \cr & \cos \theta > {\cos ^2}\theta \cr} $$

Answer: Option D

$$\eqalign{ & \cos \pi x = {x^2} - x + \frac{5}{4} \cr & = {x^2} - 2 \times x \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4} \cr & = {\left( {x - \frac{1}{2}} \right)^2} + 1 > 1 \cr & = - 1 \leqslant \cos x \leqslant 1 \cr} $$
∴ So, value of x is none of the above

Answer: Option B

$$\sin \frac{{\pi x}}{2} = {x^2} - 2x + 2$$
Put value of x from options
$$\eqalign{ & x = 1 \cr & \sin \frac{\pi }{2} \times 1 = {1^2} - 2 \times 1 + 2 \cr & \Rightarrow \sin {90^ \circ } = 1 - 2 + 2 \cr & \Rightarrow 1 = 1 \cr} $$

Answer: Option B

Angles of triangle are in ration 2 : 7 : 11
According to the question,
⇒ 2x + 7x + 11x = 180
⇒ 20x = 180
⇒ x = 9
Angles of triangle is 18°, 63°, 99°

Answer: Option B

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$

Answer: Option A

$$\eqalign{ & {\text{If }}\tan \theta = 1 \cr & {\text{It means }}\theta = {45^ \circ } \cr & = \frac{{8\sin \theta + 5\cos \theta }}{{{{\sin }^3}\theta - 2{{\cos }^3}\theta + 7\cos \theta }} \cr & = \frac{{8\sin {{45}^ \circ } + 5\cos {{45}^ \circ }}}{{{{\sin }^3}{{45}^ \circ } - 2{{\cos }^3}{{45}^ \circ } + 7\cos {{45}^ \circ }}} \cr & = \frac{{8 \times \frac{1}{{\sqrt 2 }} + 5 \times \frac{1}{{\sqrt 2 }}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} + 7\left( {\frac{1}{{\sqrt 2 }}} \right)}} \cr & = 2 \cr} $$