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

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