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Trigonometric Ratio and Identities

JEE Mathematics · 51 questions · Page 4 of 6 · Click an option or "Show Solution" to reveal answer

Q31
If 0 < x, y < π\pi and cosx + cosy −- cos(x + y) = 32{3 \over 2}, then sinx + cosy is equal to :
A 1+32{{1 + \sqrt 3 } \over 2}
B 12{{1 \over 2}}
C 32{{\sqrt 3 } \over 2}
D 1−32{{1 - \sqrt 3 } \over 2}
Correct Answer
Option A
Solution
2cos⁡(x+y2)cos⁡(x−y2)−[2cos⁡2(x+y2)−1]=322\cos \left( {{{x + y} \over 2}} \right)\cos \left( {{{x - y} \over 2}} \right) - \left[ {2{{\cos }^2}\left( {{{x + y} \over 2}} \right) - 1} \right] = {3 \over 2}
2cos⁡(x+y2)[cos⁡(x−y2)−cos⁡(x+y2)]=122\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}
2cos⁡(x+y2)[2sin⁡(x2).sin⁡(y2)]=122\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}
cos⁡(x+y2).sin⁡(x2).sin⁡(y2)=18\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}

Possible when

x2=30∘{x \over 2} = 30^\circ

&

y2=30∘{y \over 2} = 30^\circ
x=y=60∘x = y = 60^\circ
sin⁡x+cos⁡y=32+12=3+12\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}
Q32
If for x ∈\in (0,π2)\left( {0,{\pi \over 2}} \right), log10sinx + log10cosx = −-1 and log10(sinx + cosx) = 12{1 \over 2}(log10 n −- 1), n > 0, then the value of n is equal to :
A 16
B 9
C 12
D 20
Correct Answer
Option C
Solution
log⁡10sin⁡x+log⁡10cos⁡x=−1,x∈(0,π/2)log⁡10(sin⁡xcos⁡x)=−1⇒sin⁡xcos⁡x=10−1=110log⁡10(sin⁡x+cos⁡x)=12(log⁡10n−1),n>02log⁡10(sin⁡x+cos⁡x)=(log⁡10n−log⁡1010)⇒log⁡10(sin⁡x+cos⁡x)2=log⁡10(n10)⇒(sin⁡x+cos⁡x)2=n10⇒sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=n10⇒1+2(110)=n10⇒1210=n10∴n=12\begin{aligned} & \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\ & \log _{10}(\sin x \cos x)=-1 \\\\ & \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\ & \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\ & 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10} 10\right) \\\\ & \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}({n \over {10}}) \\\\ & \Rightarrow (\sin x+\cos x)^2={n \over {10}} \\\\ & \Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{n}{10} \\\\ & \Rightarrow 1+2({1 \over {10}})={n \over {10}} \Rightarrow {12 \over {10}}={n \over {10}} \\\\ & \therefore n=12 \end{aligned}
Q33
If 15sin4α\alpha + 10cos4α\alpha = 6, for some α\alpha∈\inR, then the value of 27sec6α\alpha + 8cosec6α\alpha is equal to :
A 500
B 400
C 250
D 350
Correct Answer
Option C
Solution
 Given, 15sin⁡4α+10cos⁡4α=6⇒15sin⁡4α+10cos⁡4α=6(sin⁡2α+cos⁡2α)2⇒15sin⁡4α+10cos⁡4α=6(sin⁡4α+cos⁡4α+2sin⁡2αcos⁡2α)⇒9sin⁡4α+4cos⁡4α−12sin⁡2αcos⁡2α=0⇒(3sin⁡2α−2cos⁡2α)2=0⇒3sin⁡2α−2cos⁡2α=0⇒3sin⁡2α=2cos⁡2α⇒tan⁡2α=2/3∴cot⁡2α=3/2 Now, 27sec⁡6α+8cosec⁡6α=27(sec⁡2α)3+8(cosec⁡2α)3=27(1+tan⁡2α)3+8(+cot⁡2α)3=27(1+23)3+8(1+32)3=250\begin{aligned} & \text{ Given, } 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6 \\\\ & \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2 \\\\ & \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^4 \alpha+\cos ^4 \alpha+2 \sin ^2 \alpha \cos ^2 \alpha\right) \\\\ & \Rightarrow 9 \sin ^4 \alpha+4 \cos ^4 \alpha-12 \sin ^2 \alpha \cos ^2 \alpha=0 \\\\ & \Rightarrow \quad\left(3 \sin ^2 \alpha-2 \cos ^2 \alpha\right)^2=0 \\\\ & \Rightarrow \quad 3 \sin ^2 \alpha-2 \cos ^2 \alpha=0 \\\\ & \Rightarrow \quad 3 \sin ^2 \alpha=2 \cos ^2 \alpha \\\\ & \Rightarrow \quad \tan ^2 \alpha=2 / 3 \\\\ & \therefore \quad \cot ^2 \alpha=3 / 2 \\\\ & \text{ Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=27\left(\sec ^2 \alpha\right)^3+8\left(\operatorname{cosec}^2 \alpha\right)^3 \\\\ & =27\left(1+\tan ^2 \alpha\right)^3+8\left(+\cot ^2 \alpha\right)^3 \\\\ & =27\left(1+\frac{2}{3}\right)^3+8\left(1+\frac{3}{2}\right)^3=250 \end{aligned}
Q34
If tan⁡(π9),x,tan⁡(7π18)\tan \left( {{\pi \over 9}} \right),x,\tan \left( {{{7\pi } \over {18}}} \right) are in arithmetic progression and tan⁡(π9),y,tan⁡(5π18)\tan \left( {{\pi \over 9}} \right),y,\tan \left( {{{5\pi } \over {18}}} \right) are also in arithmetic progression, then ∣x−2y∣|x - 2y| is equal to :
A 4
B 3
C 0
D 1
Correct Answer
Option C
Solution
x=12(tan⁡π9+tan⁡7π18)x = {1 \over 2}\left( {\tan {\pi \over 9} + \tan {{7\pi } \over {18}}} \right)

and

2y=tan⁡π9+tan⁡5π182y = \tan {\pi \over 9} + \tan {{5\pi } \over {18}}

If we interpret the angles in degrees (as suggested by the numbers 20, 50, and 70), we have :

x=12(tan⁡20∘+tan⁡70∘),x = \frac{1}{2} \left( \tan 20^\circ + \tan 70^\circ \right),

and

2y=tan⁡20∘+tan⁡50∘.2y = \tan 20^\circ + \tan 50^\circ.

The expression for ∣x−2y∣|x - 2y| is then :

∣x−2y∣=∣tan⁡20∘+tan⁡70∘2−(tan⁡20∘+tan⁡50∘)∣.|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ}{2} - \left( \tan 20^\circ + \tan 50^\circ \right)\right|.
∣x−2y∣=∣tan⁡20∘+tan⁡70∘−2tan⁡20∘−2tan⁡50∘2∣,|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ - 2 \tan 20^\circ - 2 \tan 50^\circ}{2}\right|,

which simplifies to :

∣x−2y∣=∣tan⁡70∘−tan⁡20∘−2tan⁡50∘2∣.|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|.

We know,

tan⁡70=tan⁡20+tan⁡501−tan⁡20tan⁡50⇒tan⁡70−tan⁡70⋅tan⁡20tan⁡50=tan⁡20+tan⁡50⇒tan⁡70−tan⁡50−tan⁡20−tan⁡50=0⇒tan⁡70−tan⁡20−2tan⁡50=0\begin{aligned} & \tan 70=\frac{\tan 20+\tan 50}{1-\tan 20 \tan 50} \\\\ &\Rightarrow \tan 70-\tan 70 \cdot \tan 20 \tan 50=\tan 20+\tan 50 \\\\ &\Rightarrow \tan 70-\tan 50-\tan 20-\tan 50=0 \\\\ &\Rightarrow \tan 70-\tan 20-2 \tan 50=0 \end{aligned}

∴\therefore

∣x−2y∣=∣tan⁡70∘−tan⁡20∘−2tan⁡50∘2∣|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|

=

∣02∣\left| {{0 \over 2}} \right|

= 0

Q35
The value of cot⁡π24\cot {\pi \over {24}} is :
A 2+3+2−6\sqrt 2 + \sqrt 3 + 2 - \sqrt 6
B 2+3+2+6\sqrt 2 + \sqrt 3 + 2 + \sqrt 6
C 2−3−2+6\sqrt 2 - \sqrt 3 - 2 + \sqrt 6
D 32−3−63\sqrt 2 - \sqrt 3 - \sqrt 6
Correct Answer
Option B
Solution
cot⁡θ=1+cos⁡2θsin⁡2θ=1+(3+122)(3−122)\cot \theta = {{1 + \cos 2\theta } \over {\sin 2\theta }} = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}
θ=π24\theta = {\pi \over {24}}
⇒cot⁡(π24)=1+(3+122)(3−122)\Rightarrow \cot \left( {{\pi \over {24}}} \right) = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}
=(22+3+1)(3−1)×(3+1)(3+1)= {{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 - 1} \right)}} \times {{\left( {\sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 + 1} \right)}}
=26+22+3+3+3+12= {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}
=6+2+3+2= \sqrt 6 + \sqrt 2 + \sqrt 3 + 2
Q36
If sin⁡θ+cos⁡θ=12\sin \theta + \cos \theta = {1 \over 2}, then 16(sin(2θ\theta) + cos(4θ\theta) + sin(6θ\theta)) is equal to :
A 23
B −-27
C −-23
D 27
Correct Answer
Option C
Solution
sin⁡θ+cos⁡θ=12\sin \theta + \cos \theta = {1 \over 2}
sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=14{\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}
sin⁡2θ=−34\sin 2\theta = - {3 \over 4}

Now :

cos⁡4θ=1−2sin⁡22θ\cos 4\theta = 1 - 2{\sin ^2}2\theta
=1−2(−34)2= 1 - 2{\left( { - {3 \over 4}} \right)^2}
=1−2×916=−18= 1 - 2 \times {9 \over {16}} = - {1 \over 8}
sin⁡6θ=3sin⁡2θ−4sin⁡32θ\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta
=(3−4sin⁡22θ).sin⁡2θ= (3 - 4{\sin ^2}2\theta ).\sin 2\theta
=[3−4(916)].(−34)= \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)
⇒[34]×(−34)=−916\Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}
16[sin⁡2θ+cos⁡4θ+sin⁡6θ]16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]

=

16(−34−18−916)=−2316\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23
Q37
The value of 2sin⁡(π8)sin⁡(2π8)sin⁡(3π8)sin⁡(5π8)sin⁡(6π8)sin⁡(7π8)2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right) is :
A 142{1 \over {4\sqrt 2 }}
B 14{1 \over 4}
C 18{1 \over 8}
D 182{1 \over {8\sqrt 2 }}
Correct Answer
Option C
Solution
2sin⁡(π8)sin⁡(2π8)sin⁡(3π8)sin⁡(5π8)sin⁡(6π8)sin⁡(7π8)2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)
2sin⁡2π8sin⁡22π8sin⁡23π82{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8}
sin⁡2π8sin⁡23π8{\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8}
sin⁡2π8cos⁡2π8{\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8}
14sin⁡2(π4)=18{1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}
Q38
α=sin⁡36∘\alpha = \sin 36^\circ is a root of which of the following equation?
A 16x4−10x2−5=016{x^4} - 10{x^2} - 5 = 0
B 16x4+20x2−5=016{x^4} + 20{x^2} - 5 = 0
C 16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 0
D 4x4−10x2+5=04{x^4} - 10{x^2} + 5 = 0
Correct Answer
Option C
Solution

Given that α=sin⁡36∘\alpha = \sin 36^\circ, we need to determine which equation it is a root of.

We start with the known relationship for cos⁡72∘\cos 72^\circ: cos⁡72∘=5−14 \cos 72^\circ = \dfrac{\sqrt{5}-1}{4} Using the double-angle formula for cosine: cos⁡72∘=1−2sin⁡236∘ \cos 72^\circ = 1 - 2 \sin^2 36^\circ Substitute α\alpha for sin⁡36∘\sin 36^\circ: 1−2α2=5−14 1 - 2\alpha^2 = \dfrac{\sqrt{5}-1}{4} Multiply both sides by 4: 4−8α2=5−1 4 - 8\alpha^2 = \sqrt{5} - 1 Add 1 to both sides: 5−8α2=5 5 - 8\alpha^2 = \sqrt{5} Square both sides to eliminate the radical: (5−8α2)2=5 (5 - 8\alpha^2)^2 = 5 Expand the left side: 25+64α4−80α2=5 25 + 64\alpha^4 - 80\alpha^2 = 5 Simplify by subtracting 5 from both sides: 64α4−80α2+20=0 64\alpha^4 - 80\alpha^2 + 20 = 0 Divide the entire equation by 4: 16α4−20α2+5=0 16\alpha^4 - 20\alpha^2 + 5 = 0 Thus, the equation 16α4−20α2+5=016\alpha^4 - 20\alpha^2 + 5 = 0 is the one for which α=sin⁡36∘\alpha = \sin 36^\circ is a root.

Q39
The value of cos⁡(2π7)+cos⁡(4π7)+cos⁡(6π7)\cos \left( {{{2\pi } \over 7}} \right) + \cos \left( {{{4\pi } \over 7}} \right) + \cos \left( {{{6\pi } \over 7}} \right) is equal to :
A −-1
B −-12{1 \over 2}
C −-13{1 \over 3}
D −-14{1 \over 4}
Correct Answer
Option B
Solution
cos⁡2π7+cos⁡4π7+cos⁡6π7=sin⁡3(π7)sin⁡π7cos⁡(2π7+6π7)2\cos {{2\pi } \over 7} + \cos {{4\pi } \over 7} + \cos {{6\pi } \over 7} = {{\sin 3\left( {{\pi \over 7}} \right)} \over {\sin {\pi \over 7}}}\cos {{\left( {{{2\pi } \over 7} + {{6\pi } \over 7}} \right)} \over 2}
=sin⁡(3π7) . cos⁡(4π7)sin⁡(π7)= {{\sin \left( {{{3\pi } \over 7}} \right)\,.\,\cos \left( {{{4\pi } \over 7}} \right)} \over {\sin \left( {{\pi \over 7}} \right)}}
=2sin⁡4π7cos⁡4π72sin⁡π7= {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}}
=sin⁡(8π7)2sin⁡π7=−sin⁡π72sin⁡π7=−12= {{\sin \left( {{{8\pi } \over 7}} \right)} \over {2\sin {\pi \over 7}}} = {{ - \sin {\pi \over 7}} \over {2\sin {\pi \over 7}}} = {{ - 1} \over 2}
Q40
16sin⁡(20∘)sin⁡(40∘)sin⁡(80∘)16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ ) is equal to :
A 3\sqrt 3
B 23\sqrt 3
C 3
D 43\sqrt 3
Correct Answer
Option B
Solution
16sin⁡20∘ . sin⁡40∘ . sin⁡80∘16\sin 20^\circ \,.\,\sin 40^\circ \,.\,\sin 80^\circ
=4sin⁡60∘= 4\sin 60^\circ

{∵\because

4sin⁡θ . sin⁡(60∘−θ) . sin⁡(60∘+θ)=sin⁡3θ4\sin \theta \,.\,\sin (60^\circ - \theta )\,.\,\sin (60^\circ + \theta ) = \sin 3\theta

}

=23= 2\sqrt 3
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