1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The value of $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$ is -
A
$${1 \over {256}}$$
B
$${1 \over {2}}$$
C
$${1 \over {1024}}$$
D
$${1 \over {512}}$$

Explanation

Given $$\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$$

Let $${\pi \over {{2^{10}}}}\, = \,\theta $$

$$ \therefore $$ $${\pi \over {{2^9}}}\, = \,2\theta $$

$${\pi \over {{2^8}}}\, = \,{2^2}\theta $$

$${\pi \over {{2^7}}}\, = \,{2^3}\theta $$
.
.

$${\pi \over {{2^2}}}\, = \,{2^8}\theta $$

So given term becomes,

$$\cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta $$$$.\sin {\pi \over {{2^{10}}}}$$

= $$(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$$

= $${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$$

= $${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$$

= $${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$$

= $${1 \over {{2^9}}}$$ = $${1 \over {512}}$$

Note :
$$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$$ = $${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Morning Slot

The maximum value of 3cos$$\theta $$ + 5sin $$\left( {\theta - {\pi \over 6}} \right)$$ for any real value of $$\theta $$ is :
A
$$\sqrt {34} $$
B
$$\sqrt {31} $$
C
$$\sqrt {19} $$
D
$${{\sqrt {79} } \over 2}$$

Explanation

y = 3cos$$\theta $$ + 5 $$\left( {\sin \theta {{\sqrt 3 } \over 2} - \cos \theta {1 \over 2}} \right)$$

$${{5\sqrt 3 } \over 2}$$ sin$$\theta $$ + $${1 \over 2}$$cos$$\theta $$

ymax = $$\sqrt {{{75} \over 4} + {1 \over 4}} $$ = $$\sqrt {19} $$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 8th April Morning Slot

If cos($$\alpha $$ + $$\beta $$) = 3/5 ,sin ( $$\alpha $$ - $$\beta $$) = 5/13 and 0 < $$\alpha , \beta$$ < $$\pi \over 4$$, then tan(2$$\alpha $$) is equal to :
A
21/16
B
63/52
C
33/52
D
63/16

Explanation

Given $$0 < \alpha < {\pi \over 4}$$

and $$0 < \beta < {\pi \over 4}$$

$$ \therefore $$ $$0 > - \beta > - {\pi \over 4}$$

$$ \therefore $$ $$0 < \alpha + \beta < {\pi \over 2}$$

and $$ - {\pi \over 4} < \alpha - \beta < {\pi \over 4}$$

As cos($$\alpha $$ + $$\beta $$) = 3/5

so $${\tan \left( {\alpha + \beta } \right) = {4 \over 3}}$$

As sin( $$\alpha $$ - $$\beta $$) = 5/13

so $${\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}$$

Now tan(2$$\alpha $$) = tan($$\alpha $$ + $$\beta $$ + $$\alpha $$ - $$\beta $$)

= $${{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$$

= $${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$$ = $${{63} \over {16}}$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th April Morning Slot

The value of cos210° – cos10°cos50° + cos250° is
A
$${3 \over 2} + \cos {20^o}$$
B
$${3 \over 4}$$
C
$${3 \over 2}(1 + \cos {20^o})$$
D
$${3 \over 2}$$

Explanation

cos210° – cos10°cos50° + cos250°

= $${1 \over 2}$$[ 2cos210° – 2cos10°cos50° + 2cos250°]

= $${1 \over 2}$$[ 1 + cos20° - cos60° - cos40° + 1 + cos100°]

= $${1 \over 2}$$[ 2 - $${1 \over 2}$$ + cos20° + cos100° - cos40°]

= $${1 \over 2}$$[ $${3 \over 2}$$ + 2cos60°cos40° - cos40°]

= $${1 \over 2}$$[ $${3 \over 2}$$ + cos40° - cos40°]

= $${3 \over 4}$$

Questions Asked from Trigonometric Functions & Equations

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