1

The sum of all values of $\theta$ $\in $$\left( {0,{\pi \over 2}} \right) satisfying sin2 2\theta + cos4 2\theta = {3 \over 4} is - A {{5\pi } \over 4} B {\pi \over 2} C \pi D {{3\pi } \over 8} ## Explanation sin22\theta + cos42\theta = {3 \over 4},$$\theta$ $\in$ $\left( {0,{\pi \over 2}} \right)$

$\Rightarrow$  1 $-$ cos22$\theta$ + cos42$\theta$ = ${3 \over 4}$

$\Rightarrow$  4cos2$\theta$ $-$ 4cos22$\theta$ + 1 = 0

$\Rightarrow$  (2cos22$\theta$ $-$ 1)2 = 0

$\Rightarrow$  cos22$\theta$ = ${1 \over 2}$ = cos2${{\pi \over 4}}$

$\Rightarrow$  2$\theta$ = n$\pi$ $\pm$ ${\pi \over 4}$, n $\in$ I

$\Rightarrow$  $\theta$ = ${{n\pi } \over 2} \pm {\pi \over 8}$

$\Rightarrow$  $\theta$ = ${\pi \over 8},{\pi \over 2} - {\pi \over 8}$

Sum of solutions ${\pi \over 2}$
2

### 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 :
$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$ = ${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$
3

### 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 = $\sqrt {{{75} \over 4} + {1 \over 4}}$ = $\sqrt {19}$
4

### 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($\alpha$ + $\beta$ + $\alpha$ - $\beta$)

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

= ${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$ = ${{63} \over {16}}$