Trigonometry - Study Mode

[#346] The minimum value of 2sin 2 θ + 3cos 2 θ is ?
Correct Answer

(C) 2

Explanation

Solution: Let x = 2sin 2 θ + 3cos 2 θ ⇒ x = 2sin 2 θ + 2cos 2 θ + cos 2 θ ⇒ x = 2(sin 2 θ + cos 2 θ) + cos 2 θ ⇒ x = 2 + cos 2 θ xa0 xa0 [since sin 2 θ + cos 2 θ = 1] Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2 Alternative Solution: 2sin 2 θ + 3cos 2 θ Minimum value is 2, [If x sin 2 θ + y cos 2 θ,
If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]

[#347] Maximum value of (2sinθ + 3cosθ) is?
Correct Answer

(B) $$sqrt {13} $$

Explanation

Solution: $$eqalign{
& left( {2sin heta + 3cos heta }
ight) cr
& { ext{Maximum value of}} cr
& asin heta + bcos heta cr
& = sqrt {{a^2} + {b^2}} cr
& = sqrt {{2^2} + {3^2}} cr
& = sqrt {4 + 9} cr
& = sqrt {13} cr} $$

[#348] The equation $${cos ^2} heta $$xa0 = $$frac{{{{left( {x + y}
ight)}^2}}}{{4xy}}$$ xa0 is only possible when ?
Correct Answer

(C) x = y

Explanation

Solution: $$eqalign{
& {cos ^2} heta = frac{{{{left( {x + y}
ight)}^2}}}{{4xy}} cr
& { ext{Max value of }}{cos ^2} heta = 1 cr
& Rightarrow 1 = frac{{{{left( {x + y}
ight)}^2}}}{{4xy}} cr
& Rightarrow 4xy = {left( {x + y}
ight)^2} cr
& Rightarrow 4xy = {x^2} + {y^2} + 2xy cr
& Rightarrow 0 = {x^2} + {y^2} - 2xy cr
& Rightarrow 0 = {left( {x - y}
ight)^2} cr
& Rightarrow 0 = x - y cr
& Rightarrow x = y cr} $$

[#349] The greatest value of sin 4 θ + cos 4 θ is?
Correct Answer

(D) 1

Explanation

Solution: $$eqalign{
& {sin ^2} heta + {cos ^2} heta = 1 cr
& { ext{Squaring both sides}} cr
& {sin ^4} heta + {cos ^4} heta cr
& = 1 - 2{sin ^2} heta . {cos ^2} heta cr
& { ext{Put}}, heta = {90^ circ } cr
& = 1 - 2{sin ^2}{90^ circ } imes {cos ^2}{90^ circ } cr
& = 1 - 0 cr
& = 1 cr} $$

[#350] Which one of the following is true for 0° < θ < 90° ?
Correct Answer

(C) cosθ > cos 2 θ

Explanation

Solution: $$eqalign{
& { ext{Put, }} heta = {60^ circ } cr
& Rightarrow cos heta > {cos ^2} heta cr
& Rightarrow cos {60^ circ } > {cos ^2}{60^ circ } cr
& Rightarrow frac{1}{2} > frac{1}{4} cr
& cos heta > {cos ^2} heta cr} $$