1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

For any $$\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)$$, the expression

$$3{(\cos \theta - \sin \theta )^4}$$$$ + 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta $$

equals :
A
13 – 4 cos2$$\theta $$ + 6sin2$$\theta $$cos2$$\theta $$
B
13 – 4 cos6$$\theta $$
C
13 – 4 cos2$$\theta $$ + 6cos2$$\theta $$
D
13 – 4 cos4$$\theta $$ + 2sin2$$\theta $$cos2$$\theta $$

Explanation

Given,

3(sin$$\theta $$ $$-$$ cos$$\theta $$)4 + 6(sin$$\theta $$ + cos$$\theta $$)2 + 4sin6$$\theta $$

= 3[(sin$$\theta $$ $$-$$ cos$$\theta $$)2]2 + 6 (sin2$$\theta $$ + cos2$$\theta $$ + 2sin$$\theta $$cos$$\theta $$) + 4sin6$$\theta $$

= 3[sin2$$\theta $$ + cos2$$\theta $$ $$-$$2sin$$\theta $$cos$$\theta $$]2 + 6(1 + sin2$$\theta $$) + 4sin6$$\theta $$

= 3(1 $$-$$ sin2$$\theta $$)2 + 6(1 + sin2$$\theta $$) + 4sin6$$\theta $$

= 3 (1 $$-$$ 2 sin2$$\theta $$ + sin22$$\theta $$) + 6 + 6sin2$$\theta $$ + 4sin6$$\theta $$

= 3 $$-$$ 6sin2$$\theta $$ + 3sin22$$\theta $$ + 6 + 6sin2$$\theta $$ + 4sin6$$\theta $$

= 9 + 3sin22$$\theta $$ + 4 sin6$$\theta $$

= 9 + 3(2sin$$\theta $$cos$$\theta $$)2 + 4(1 $$-$$ cos2$$\theta $$)3

= 9 + 12sin2$$\theta $$ cos2$$\theta $$ + 4 (1 $$-$$ cos6$$\theta $$ $$-$$ 3cos2$$\theta $$ + 3cos4$$\theta $$)

= 13 + 12 (1 $$-$$ cos2$$\theta $$ $$-$$ 4cos6$$\theta $$ $$-$$ 12cos$$\theta $$ + 12 cos4$$\theta $$

= 13 + 12 cos2$$\theta $$ $$-$$ 12 cos4$$\theta $$ $$-$$ 4cos6$$\theta $$ $$-$$ 12 cos2$$\theta $$ + 12 cos4$$\theta $$

= 13 $$-$$ 4 cos6$$\theta $$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

If  0 $$ \le $$ x < $${\pi \over 2}$$,  then the number of values of x for which sin x $$-$$ sin 2x + sin 3x = 0, is :
A
3
B
1
C
4
D
2

Explanation

sin x $$-$$ sin 2x + sin 3x = 0        $$x \in \left[ {0,{\pi \over 2}} \right)$$

$$ \Rightarrow $$  (sin3x + sinx) $$-$$ sin2x = 0

$$ \Rightarrow $$  2sin2x.cos2x $$-$$ sin2x = 0

$$ \Rightarrow $$  sin2x (2cosx $$-$$ 1) = 0

sin 2x = 0

x = 0

and cos x = $${1 \over 2}$$

and x = $${\pi \over 3}$$

two solutions
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

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 $${\pi \over 2}$$
4
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 :
$$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$$ = $${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$$

Questions Asked from Trigonometric Functions & Equations

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