Trigonometric Ratio and Identities — Page 3 | JEE Mathematics

Q21

Let

\alpha ,\,\beta

be such that

\pi < \alpha - \beta < 3\pi

. If

sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}

and

\cos \alpha + \cos \beta = - {{27} \over {65}}

then the value of

\cos {{\alpha - \beta } \over 2}

:

A

{{ - 6} \over {65}}\,\,

B

{3 \over {\sqrt {130} }}

C

{6 \over {65}}

D

- {3 \over {\sqrt {130} }}

Correct Answer

Option D

Solution

Given

sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}

.........(1) and

\cos \alpha + \cos \beta = - {{27} \over {65}}

........(2) Square and add (1) and (2) you will get

2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)

= {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}

$\Rightarrow$

2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}

$\Rightarrow$

4{\cos ^2}{{\alpha - \beta } \over 2}

= {{1170} \over {{{\left( {65} \right)}^2}}}

$\Rightarrow$

{\cos ^2}{{\alpha - \beta } \over 2}

= {9 \over {130}}

$\therefore$

\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}

[ But

\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}

as

\pi < \alpha - \beta < 3\pi

$\Rightarrow$

{\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}

$\Rightarrow$

\cos {{\alpha - \beta } \over 2} < 0

] So

\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}

Q22

If

0 < x < \pi

and

\cos x + \sin x = {1 \over 2},

then

\tan x

is :

A

{{\left( {1 - \sqrt 7 } \right)} \over 4}

B

{{\left( {4 - \sqrt 7 } \right)} \over 3}

C

- {{\left( {4 + \sqrt 7 } \right)} \over 3}

D

{{\left( {1 + \sqrt 7 } \right)} \over 4}

Correct Answer

Option C

Solution

\cos x + \sin x = {1 \over 2}

\Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}

\Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}

\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]

\Rightarrow 1 + \sin 2x = {1 \over 4}

\Rightarrow \sin 2x = - {3 \over 4},

so

x

is obtuse and

{{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}

\Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0

$\therefore$

\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}

= {{ - 4 \pm \sqrt 7 } \over 3}

as

\tan x < 0\,

$\therefore$

\tan x = {{ - 4 - \sqrt 7 } \over 3}

Q23

Let

\cos \left( {\alpha + \beta } \right) = {4 \over 5}

and

\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},

where

0 \le \alpha ,\,\beta \le {\pi \over 4}.

Then

tan\,2\alpha

=

A

{56 \over 33}

B

{19 \over 12}

C

{20 \over 7}

D

{25 \over 16}

Correct Answer

Option A

Solution

\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}

\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}

\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]

= {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}

Q24

If

A = {\sin ^2}x + {\cos ^4}x,

then for all real

x

:

A

{{13} \over {16}} \le A \le 1

B

1 \le A \le 2

C

{3 \over 4} \le A \le {{13} \over {16}}

D

{{3} \over {4}} \le A \le 1

Correct Answer

Option D

Solution

A = {\sin ^2}x + {\cos ^4}x

= {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)

= {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}

= 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)

Now

0 \le {\sin ^2}\left( {2x} \right) \le 1

\Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}

\Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}

\Rightarrow 1 \ge A \ge {3 \over 4}

Q25

The expression

{{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}

can be written as:

A

\sin {\rm A}\,\cos {\rm A} + 1

B

\,\sec {\rm A}\,\cos ec{\rm A} + 1

C

\tan {\rm A} + \cot {\rm A}

D

\sec {\rm A} + \cos ec{\rm A}

Correct Answer

Option B

Solution

Given expression can be written as

{{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}

(As

\tan A = {{\sin A} \over {\cos A}}

and

\cot A = {{\cos A} \over {\sin A}}

)

= {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}

= {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}

= 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A

Q26

Let

f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)

where

x \in R

and

k \ge \,1.

Then

{f_4}\left( x \right) - {f_6}\left( x \right)\,\,

equals :

A

{1 \over 4}

B

{1 \over 12}

C

{1 \over 6}

D

{1 \over 3}

Correct Answer

Option B

Solution

Let

{f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)

Consider

{f_4}\left( x \right) - {f_6}\left( x \right)

$=$

{1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)

= {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]

= {1 \over 4} - {1 \over 6} = {1 \over {12}}

Q27

If

5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9

, then the value of

\cos 4x

is :

A

{1 \over 3}

B

{2 \over 9}

C

- {7 \over 9}

D

- {3 \over 5}

Correct Answer

Option C

Solution

Given that,

5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9

\Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9

Let

{\cos ^2}x = t,

then we have

5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9

\Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9

\Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t

\Rightarrow 9{t^2} + 12t - 5 = 0

\Rightarrow 9{t^2} + 15t - 3t - 5 = 0

\Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0

\Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0

$\therefore$

t = {1 \over 3}

and

t = - {5 \over 3}

If

t = - {5 \over 3}

then

{\cos ^2}x

is negative So ,

t

can not be

- {5 \over 3}

. So, correct value of

t = {1 \over 3}

then

\cos {}^2x = t = {1 \over 3}

\therefore\,\,\,

\cos 4x

= 2{\cos ^2}2x - 1

= 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1

= 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1

= 2.{\left( { - {1 \over 3}} \right)^2} - 1

= {2 \over 9} - 1

= - {7 \over 9}

Q28

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

Correct Answer

Option B

Solution

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 $\theta$

Q29

The value of

{\cos ^3}\left( {{\pi \over 8}} \right)

{\cos}\left( {{3\pi \over 8}} \right)

+

{\sin ^3}\left( {{\pi \over 8}} \right)

{\sin}\left( {{3\pi \over 8}} \right)

is :

A

{1 \over {\sqrt 2 }}

B

{1 \over 2}

C

{1 \over 4}

D

{1 \over 2{\sqrt 2 }}

Correct Answer

Option D

Solution

{\cos ^3}\left( {{\pi \over 8}} \right)

{\cos}\left( {{3\pi \over 8}} \right)

+

{\sin ^3}\left( {{\pi \over 8}} \right)

{\sin}\left( {{3\pi \over 8}} \right)

=

{\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)

=

\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)\left[ {{{\cos }^2}\left( {{\pi \over 8}} \right) + {{\sin }^2}\left( {{\pi \over 8}} \right)} \right]

=

\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)

$\times$ 1 =

{1 \over 2} \times 2\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)

=

{1 \over 2}\sin \left( {{\pi \over 4}} \right)

=

{1 \over {2\sqrt 2 }}

Q30

If L = sin2

\left( {{\pi \over {16}}} \right)

- sin2

\left( {{\pi \over {8}}} \right)

and M = cos2

\left( {{\pi \over {16}}} \right)

- sin2

\left( {{\pi \over {8}}} \right)

, then :

A L =

- {1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}

B M =

{1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}

C M =

{1 \over {4\sqrt 2 }} + {1 \over 4}\cos {\pi \over 8}

D L =

{1 \over {4\sqrt 2 }} - {1 \over 4}\cos {\pi \over 8}

Correct Answer

Option B

Solution

We will use here those two formulas, sin2 $\theta$ =

{{1 - \cos 2\theta } \over 2}

and cos2 $\theta$ =

{{1 + \cos 2\theta } \over 2}

L = sin2

\left( {{\pi \over {16}}} \right)

- sin2

\left( {{\pi \over {8}}} \right)

$\Rightarrow$ L =

\left( {{{1 - \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)

-

\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)

$\Rightarrow$ L =

{1 \over 2}\left( {\cos \left( {{\pi \over 4}} \right) - \cos \left( {{\pi \over 8}} \right)} \right)

$\Rightarrow$ L =

{1 \over {2\sqrt 2 }} - {1 \over 2}\cos \left( {{\pi \over 8}} \right)

M = cos2

\left( {{\pi \over {16}}} \right)

- sin2

\left( {{\pi \over {8}}} \right)

$\Rightarrow$ M =

\left( {{{1 + \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)

-

\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)

$\Rightarrow$ M =

{1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}