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JEE • MATHEMATICS • MULTIPLE AND SUBMULTIPLE ANGLES

Multiple and submultiple angles practice · Mixed difficulty

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Showing 91-120 of 122 questions
91 Easy
91 of 122
Analyzing +4 | -1
tan⁡  α+2  tan⁡  2α+4  tan⁡  4α+8  cot⁡  8α\tan \,\,\alpha + 2\,\,\tan \,\,2\alpha + 4\,\,\tan \,\,4\alpha + 8\,\,\cot \,\,8\alpha is equal to
View answer and explanation
Correct option: C
cot⁡  α\cot \,\,\alpha
Option C (cot⁡α\cot \alpha) is correct because the given expression simplifies to cot⁡α\cot \alpha using the identity cot⁡2x=cot⁡x−tan⁡x2\cot 2x = \frac{{\cot x - \tan x}}{2}.
92 Medium
92 of 122
Application +4 | -1
The value of 16cos⁡(2π15)⋅cos⁡(4π15)cos⁡(8π15)⋅cos⁡(16π15)16 \cos \left(\frac{2 \pi}{15}\right) \cdot \cos \left(\frac{4 \pi}{15}\right) \cos \left(\frac{8 \pi}{15}\right) \cdot \cos \left(\frac{16 \pi}{15}\right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression can be simplified using the identity 2sin⁡xcos⁡x=sin⁡2x2 \sin x \cos x = \sin 2x repeatedly.
93 Medium
93 of 122
Application +4 | -1
The minimum and maximum values of absin⁡x+b(1−a2)cos⁡x+c(∣a∣<1,b>0)a b \sin x+b \sqrt{\left(1-a^2\right)} \cos x+c(|a|<1, b>0) respectively are
View answer and explanation
Correct option: C
{c−b,b+c}\{c-b, b+c\}
Option C ({c−b,b+cc-b, b+c}) is correct because by rewriting the expression and using trigonometric identities, the minimum and maximum values can be found.
94 Medium
94 of 122
Application +4 | -1
If 2cos⁡θ+sin⁡θ=12 \cos \theta+\sin \theta=1, then 7cos⁡θ+6sin⁡θ7 \cos \theta+6 \sin \theta equals
View answer and explanation
Correct option: D
2 or 6
Option D (2 or 6) is correct because by using trigonometric substitutions and solving the resulting quadratic equation, we find two possible values for the expression 7cos⁡θ+6sin⁡θ7 \cos \theta+6 \sin \theta.
95 Easy
95 of 122
Application +4 | -1
Suppose 0<t<π0 < t < \pi and sin⁡  t+cos⁡  t=15\sin \,\,t + \cos \,\,t = \frac{1}{5}. Then, tan⁡  t2\tan \,\,\frac{t}{2} is equal to
View answer and explanation
Correct option: A
2
Option A, 2, is correct because it is the value of tan⁡t2\tan \frac{t}{2} obtained by substituting the given expression into the trigonometric identity and solving the resulting quadratic equation.
96 Medium
96 of 122
Application +4 | -1
If sin⁡β\sin \beta is the geometric mean between sin⁡α\sin \alpha and cos⁡α\cos \alpha, then cos⁡2β\cos 2 \beta is equal to
View answer and explanation
Correct option: A
2sin⁡2(π4−α)2 \sin ^2\left(\frac{\pi}{4}-\alpha\right)
Option A (2sin⁡2(π4−α)2 \sin ^2\left(\frac{{\pi}}{{4}}-\alpha\right)) is correct because by using the given geometric mean condition and trigonometric identities, we can express cos⁡2β\cos 2\beta in terms of α\alpha and simplify it to the given form.
97 Easy
97 of 122
Application +4 | -1
sin⁡4  π8+sin⁡4  3 π8+sin⁡4  5 π8+sin⁡4  7π8  {\sin ^4}\,\,\frac{\pi }{8} + {\sin ^4}\,\,\frac{{3\,\pi }}{8} + {\sin ^4}\,\,\frac{{5\,\pi }}{8} + {\sin ^4}\,\,\frac{{7\pi }}{8}\; is equal to
View answer and explanation
Correct option: B
3 / 2\mathrm{3}\,\mathrm{/}\,\mathrm{2}
Option B (3/2) is correct because the given trigonometric expression simplifies to 3/2 using trigonometric identities and properties.
98 Medium
98 of 122
Application +4 | -1
The value of cos⁡Acos⁡2Acos⁡22Acos⁡23A…cos⁡2n−1A\cos A \cos 2 A \cos 2^2 A \cos 2^3 A \ldots \cos 2^{n-1} A is
View answer and explanation
Correct option: B
sin⁡2nA2nsin⁡A\frac{\sin 2^n A}{2^n \sin A}
Option B is correct because the given expression can be simplified using the identity 2sin⁡Acos⁡A=sin⁡2A2 \sin A \cos A = \sin 2A repeatedly.
99 Easy
99 of 122
Application +4 | -1
The value of cos⁡π22⋅cos⁡π23.......cos⁡π210⋅sin⁡π210\cos \frac{\pi }{{{2^2}}} \cdot \cos \frac{\pi }{{{2^3}}}.......\cos \frac{\pi }{{{2^{10}}}} \cdot \sin \frac{\pi }{{{2^{10}}}} is
View answer and explanation
Correct option: C
1512\frac{1}{{512}}
Option C (1512\frac{1}{{512}}) is correct because it is the simplified result after applying the trigonometric identity for the product of cosines and sine.
100 Easy
100 of 122
Application +4 | -1
If cos⁡  (x−y),  cos⁡x\cos \,\,\left( {x - y} \right),\,\,\cos x and cos⁡  (x+y)\cos \,\,\left( {x + y} \right) are in H.P., then ∣ cos⁡x  sec⁡  y2 ∣\left| {\,\cos x\,\,\sec \,\,\frac{y}{2}\,} \right| equals
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because after applying the properties of Harmonic Progression and trigonometric identities, the expression simplifies to 2\sqrt{2}.
101 Medium
101 of 122
Analyzing +4 | -1
If cos⁡A+cos⁡B+cos⁡C=0\cos A + \cos B + \cos C = 0, then cos⁡  3A+cos⁡  3B+cos⁡  3C\cos \,\,3A + \cos \,\,3B + \cos \,\,3C is equal to
View answer and explanation
Correct option: B
12  cos⁡A  cos⁡B  cos⁡C12\,\,\cos A\,\,\cos B\,\,\cos C
Option B (12,cos⁡A,cos⁡B,cos⁡C12\\,\cos A\\,\cos B\\,\cos C) is the correct answer because it follows from trigonometric identities and the given condition cos⁡A+cos⁡B+cos⁡C=0\cos A + \cos B + \cos C = 0.
102 Medium
102 of 122
Application +4 | -1
∑r=1n−1  cos⁡2  r  πn\sum \limits_{r = 1}^{n - 1} \,\,{\cos ^2}\,\,\frac{{r\;\pi }}{n} is equal to
View answer and explanation
Correct option: C
n2−1\frac{n}{2} - 1
Option C (n2−1\frac{n}{2} - 1) is correct because the summation of the cosine squared terms simplifies to this expression using trigonometric identities and series summation.
103 Easy
103 of 122
Analyzing +4 | -1
sin⁡23 Asin⁡2 A−cos⁡23 Acos⁡2A=\frac{{{{\sin }^2}3\,A}}{{{{\sin }^2}\,A}} - \frac{{{{\cos }^2}3\,A}}{{{{\cos }^2}A}} =
View answer and explanation
Correct option: B
8  cos⁡  2A8\;\cos \,\,2A
Option B (8  cos⁡  2A8\;\cos \;2A) is correct because the expression simplifies to 8cos⁡2A8\cos 2A using trigonometric identities.
104 Easy
104 of 122
Application +4 | -1
If sin⁡A=336625\sin A = \frac{{336}}{{625}} where 450∘  <A<540∘450^\circ \,\, < A < 540^\circ, then sin⁡  A4=\sin \,\,\frac{A}{4} =
View answer and explanation
Correct option: C
45\frac{4}{5}
Option C (45\frac{4}{5}) is correct because given the range of A and the value of sin⁡A\sin A, we can find sin⁡A4\sin \frac{A}{4} using half-angle formulas.
105 Medium
105 of 122
Application +4 | -1
The smallest values of xx and yy satisfying x−y=π4x-y=\frac{\pi}{4} and cot⁡x+cot⁡y=2\cot x+\cot y=2 are -
View answer and explanation
Correct option: B
x=5π12,y=π6x=\frac{5 \pi}{12}, y=\frac{\pi}{6}
Option B (x=5π12,y=π6x=\frac{5 \pi}{12}, y=\frac{\pi}{6}) is correct because these values satisfy both the given equations: x−y=π4x-y=\frac{\pi}{4} and cot⁡x+cot⁡y=2\cot x+\cot y=2.

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106 Easy
106 of 122
Analyzing +4 | -1
If cos⁡  x=tan⁡y,  cos⁡  y=tan⁡  z,  cos⁡  z=tan⁡ x\cos \,\,x = \tan y,\,\,\cos \,\,y = \tan \,\,z,\,\,\cos \,\,z = \tan \,x, then the value of sin⁡x\sin x is
View answer and explanation
Correct option: D
2sin⁡ 18∘2\sin \,18^\circ
Option D (2sin⁡ 18∘2\sin \,18^\circ) is the correct answer because, after simplification and solving the trigonometric equations, the value of sin⁡x\sin x is found to be equal to 2sin⁡ 18∘2\sin \,18^\circ.
107 Easy
107 of 122
Application +4 | -1
2sin⁡(π22)sin⁡(3π22)sin⁡(5π22)sin⁡(7π22)sin⁡(9π22)2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right) is equal to:
View answer and explanation
Correct option: B
116\frac{1}{{16}}
Option B (5) is correct because setting up the integral for the area and solving for 'a' yields the odd natural number 5.
108 Easy
108 of 122
Application +4 | -1
The value of sin⁡  36∘  sin⁡  72∘  sin⁡  108∘  sin⁡144∘\sin \,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin 144^\circ is
View answer and explanation
Correct option: D
516\frac{5}{{16}}
Option D (5/16) is correct because it is the result of simplifying the given expression using trigonometric identities and known values of sine functions.
109 Easy
109 of 122
Analyzing +4 | -1
If (a−b)  sin⁡  (θ+)=(a+b)sin⁡  (θ−ϕ)   and   a  tan⁡  θ2 −  b  tan⁡  ϕ2=c\left( {a - b} \right)\,\,\sin \,\,\left( {\mathrm{\theta } + } \right) = \left( {a + b} \right)\sin \,\,(\mathrm{\theta } - \phi )\,\;\mathrm{and}\,\mathrm{\;}a\,\,\tan \,\,\frac{\mathrm{\theta }}{2}\, - \,\,b\;\tan \,\,\frac{\phi }{2} = c, then the value of sin⁡  ϕ\sin \,\,\phi is equal to
View answer and explanation
Correct option: B
2 bca2−  b2−  c2\frac{{2\,bc}}{{{a^2} - \,\,{b^2} - \,\,{c^2}}}
Option C (2 bca2− b2+ c2\frac{{2\,bc}}{{{{a^2} - \,{b^2} + \,{c^2}}}}) is correct because after simplification and substitution using the given equations, it leads to the correct expression for sin(ϕ\phi).
110 Easy
110 of 122
Application +4 | -1
If tan⁡A=1−cos⁡  Bsin⁡  B\tan A = \frac{{1 - \cos \,\,B}}{{\sin \,\,B}}, then
View answer and explanation
Correct option: A
tan⁡  2A=tan⁡B\tan \,\,2A = \tan B
Option A is correct because it correctly uses trigonometric identities to simplify the given equation and derive the relationship between tan⁡2A\tan 2A and tan⁡B\tan B.
111 Medium
111 of 122
Analyzing +4 | -1
If a=tan⁡  27θ−tan⁡  θa = \tan \,\,27\theta - \tan \,\,\theta and b=sin⁡  θcos⁡  3θ+sin⁡  3θcos⁡  9θ+sin⁡  9θcos⁡  27θb = \frac{{\sin \,\,\theta }}{{\cos \,\,3\theta }} + \frac{{\sin \,\,3\theta }}{{\cos \,\,9\theta }} + \frac{{\sin \,\,9\theta }}{{\cos \,\,27\theta }} , then
View answer and explanation
Correct option: B
a=2ba = 2b
Option B (a=2ba = 2b) is the correct answer because, after simplification and trigonometric manipulations, the expression for aa can be shown to be twice the expression for bb.
112 Easy
112 of 122
Application +4 | -1
The value of cot⁡  36∘  cot⁡  72∘\cot \,\,36^\circ \,\,\cot \,\,72^\circ is
View answer and explanation
Correct option: B
15\frac{1}{{\sqrt 5 }}
Option B (15\frac{1}{{\sqrt 5 }}) is correct because by using trigonometric identities and known values of trigonometric functions for specific angles, the expression simplifies to 15\frac{1}{{\sqrt 5 }}.
113 Medium
113 of 122
Analyzing +4 | -1
For any angle θ\theta, the expression 2cos⁡  8θ+12cos⁡  θ+1\frac{{2\cos \,\,8\theta + 1}}{{2\cos \,\,\theta + 1}} is equal to
View answer and explanation
Correct option: C
(2  cos⁡  θ−1)  (2  cos⁡  2θ−1)  (2  cos⁡  4θ−1)\left( {2\,\,\cos \,\,\theta - 1} \right)\,\,\left( {2\,\,\cos \,\,2\theta - 1} \right)\,\,\left( {2\,\,\cos \,\,4\theta - 1} \right)
Option C ((2 cos⁡ θ−1) (2 cos⁡ 2θ−1) (2 cos⁡ 4θ−1)\left( {2\,\cos \,\theta - 1} \right)\,\left( {2\,\cos \,2\theta - 1} \right)\,\left( {2\,\cos \,4\theta - 1} \right)) is correct because it is obtained by repeatedly applying the double angle formula for cosine and simplifying the expression.
114 Medium
114 of 122
Application +4 | -1
If 4n  α=π4n\;\alpha = \pi, then the value of tan⁡  α  tan⁡  2α  tan⁡  3α  tan⁡  4α…tan⁡(2n−2)α  tan⁡(2n−1)α\tan \,\,\alpha \,\,\tan \,\,2\alpha \,\,\tan \,\,3\alpha \,\,\tan \,\,4\alpha \ldots \tan \left( {2n - 2} \right)\alpha \,\,\tan \left( {2n - 1} \right)\alpha, is
View answer and explanation
Correct option: B
1
Option B is correct because the product telescopes to 1 due to the complementary angle relationship between tangent functions.
115 Hard
115 of 122
Analyzing +4 | -1
The maximum value of (cos⁡  α1)  (cos⁡  α2)  …  (cos⁡  αn)    (\cos \,\,{\alpha _1})\;(\cos \,\,{\alpha _2})\,\, \ldots \,\,(\cos \,\,{\alpha _n})\;\; under the restriction
0  ≤  α1,α2,…,αn  ≤  π20\,\, \le \,\,{\alpha _{1,}}{\alpha _2}, \ldots ,{\alpha _n}\,\, \le \,\,\frac{\pi }{2} and (cos⁡  α1) (cos⁡  α2)…(cos⁡  αn)=1    (\cos \,\,{\alpha _1})\,(\cos \,\,{\alpha _2}) \ldots (\cos \,\,{\alpha _n}) = 1\;\;, is

View answer and explanation
Correct option: A
12n/2\frac{1}{{{2^{n/2}}}}
Option A (12n/2\frac{1}{{{2^{n/2}}}}) is correct because the maximum value is achieved when all angles are equal, leading to the given result using the AM-GM inequality or similar arguments.
116 Medium
116 of 122
Application +4 | -1
If cos⁡(α+β)=0\cos (\alpha+\beta)=0 then cosec⁡(α−β)=\operatorname{cosec}(\alpha-\beta)=
View answer and explanation
Correct option: D
±sec⁡2β\pm \sec 2 \beta
Option D (±sec⁡2β\pm \sec 2 \beta) is correct because from cos⁡(α+β)=0\cos(\alpha + \beta) = 0, we can derive that α+β=(2n+1)π2\alpha + \beta = (2n+1)\frac{\pi}{2}, and then find cosec⁡(α−β)\operatorname{cosec}(\alpha - \beta) in terms of β\beta.
117 Medium
117 of 122
Analyzing +4 | -1
Value of cos⁡2(A−B)  +  cos⁡2B  −  2cos⁡(A−B)cos⁡A  cos⁡B{\cos ^2}\left( {A - B} \right)\,\, + \,\,{\cos ^2}B\,\, - \,\,2\cos \left( {A - B} \right)\cos A\,\,\cos B is
View answer and explanation
Correct option: B
sin⁡2A{\sin ^2}A
Option B (sin⁡2A{\sin ^2}A) is correct because by applying trigonometric identities and simplification, the expression reduces to sin⁡2A{\sin ^2}A.
118 Easy
118 of 122
Application +4 | -1
The value of 16  sin⁡  144∘  sin⁡  108∘  sin⁡  72∘  sin⁡  36∘16\,\,\sin \,\,144^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,36^\circ is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because by using trigonometric identities and simplifying the expression 16 sin⁡ 144∘ sin⁡ 108∘ sin⁡ 72∘ sin⁡ 36∘16\,\sin \,144^\circ \,\sin \,108^\circ \,\sin \,72^\circ \,\sin \,36^\circ, we find that it is equal to 5.
119 Medium
119 of 122
Application +4 | -1
If 3sin⁡θ+5cos⁡θ=53 \sin \theta+5 \cos \theta=5, then the value of 5sin⁡θ−3cos⁡θ5 \sin \theta-3 \cos \theta is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because by using trigonometric identities and the given equation, the value of 5sin⁡θ−3cos⁡θ5 \sin \theta - 3 \cos \theta can be derived to be 3.
120 Medium
120 of 122
Application +4 | -1
If cos⁡2x+cos⁡4x=1\cos ^2 x+\cos ^4 x=1, then the value of tan⁡4x+cot⁡4x+tan⁡2x−cot⁡2x\tan ^4 x+\cot ^4 x+\tan ^2 x-\cot ^2 x is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the given equation and trigonometric identities, the expression simplifies to 2.
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