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JEE • MATHEMATICS • MULTIPLE AND SUBMULTIPLE ANGLES

Multiple and submultiple angles practice · Medium

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Showing 1-30 of 52 questions
1 Medium
1 of 52
Application +4 | -1
The value of sin⁡12∘⋅sin⁡48∘⋅sin⁡54∘=\sin 12^\circ \cdot \sin 48^\circ \cdot \sin 54^\circ =
View answer and explanation
Correct option: A
1/81/8
Option A (1/8) is correct because by using the product-to-sum trigonometric identities, the expression simplifies to 1/8.
2 Medium
2 of 52
Application +4 | -1
If α,    β(α≠β)\alpha ,\;\,\,\beta \left( {\alpha \ne \beta } \right) satisfies the equation a a  cos⁡  θ+b  sin⁡  θ=ca\,\,\cos \,\,\theta + b\,\,\sin \,\,\theta = c, then the value of tan⁡  (α+β2)\tan \,\,\left( {\frac{{\alpha + \beta }}{2}} \right) is

View answer and explanation
Correct option: A
b/ab/a
Option A (b/a) is correct because by using trigonometric identities and subtracting the equations for α and β, we can derive the value of tan((α + β)/2).
3 Medium
3 of 52
Application +4 | -1
cos⁡4θ−sin⁡4θ{\cos ^4}\theta - {\sin ^4}\theta is equal to
View answer and explanation
Correct option: B
2  cos⁡2θ−12\,\,{\cos ^2}\theta - 1
Option B is correct because the expression simplifies to cos⁡2θ\cos 2\theta, which is equivalent to 2cos⁡2θ−12{\cos ^2}\theta - 1.
4 Medium
4 of 52
Analyzing +4 | -1
The value of
cos⁡210∘−cos⁡10∘cos⁡50∘+cos⁡250∘\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ} is :
View answer and explanation
Correct option: B
3/43 / 4
Option B (3/4) is correct because the trigonometric expression simplifies to 3/4 after applying trigonometric identities and simplifications.
5 Medium
5 of 52
Analyzing +4 | -1
In a △PQR,∠R=π2\triangle P Q R, \angle R=\frac{\pi}{2}. If tan⁡(P2)\tan \left(\frac{P}{2}\right) and tan⁡(Q2)\tan \left(\frac{Q}{2}\right) are the roots of ax2+bx+c=0,a≠0a x^2+b x+c=0, a \neq 0, then
View answer and explanation
Correct option: C
c=a+bc=a+b
Option C (c=a+bc=a+b) is correct because it is derived from the relationship between the tangent of half angles and the sum of angles in a right-angled triangle.
6 Medium
6 of 52
Application +4 | -1
If 0<α<π/60<\alpha<\pi / 6 and sin⁡α+cos⁡α=7/2\sin \alpha+\cos \alpha=\sqrt{7} / 2, then tan⁡α/2\tan \alpha / 2 is equal to
View answer and explanation
Correct option: A
7−23\frac{\sqrt{7}-2}{3}
Option A (7−23\frac{{\sqrt{{7}}-2}}{{3}}) is correct because solving the trigonometric equation and considering the given range for α\alpha, the value of tan⁡α2\tan \frac{{\alpha}}{{2}} is 7−23\frac{{\sqrt{{7}}-2}}{{3}}.
7 Medium
7 of 52
Analyzing +4 | -1
If tan⁡3Atan⁡A=k\frac{\tan ^3 A}{\tan A}=k, then sin⁡3Asin⁡A\frac{\sin 3 A}{\sin A} is equal to
View answer and explanation
Correct option: A
2kk−1\frac{2 k}{k-1}
Option A (2kk−1\frac{2 k}{k-1}) is correct because by using trigonometric identities and simplifying the given expression, we can find that sin⁡3Asin⁡A=2kk−1\frac{\sin 3 A}{\sin A} = \frac{2k}{k-1}.
8 Medium
8 of 52
Application +4 | -1
If cos⁡θa=sin⁡θb\frac{\cos \theta}{a}=\frac{\sin \theta}{b}, then asec⁡2θ+bcosec⁡2θ\frac{a}{\sec 2 \theta}+\frac{b}{\operatorname{cosec} 2 \theta} is equal to
View answer and explanation
Correct option: A
aa
Option A (aa) is correct because the expression simplifies to acos⁡2θ+bsin⁡2θ=aa \cos 2\theta + b \sin 2\theta = a.
9 Medium
9 of 52
Analyzing +4 | -1
If 5(tan⁡2x−cos⁡2x)=2cos⁡2x+95\left(\tan ^{2} x-\cos ^{2} x\right)=2 \cos 2 x+9, then the value of cos⁡4x\cos 4 x is
View answer and explanation
Correct option: A
−79-\frac{7}{9}
Option A (-7/9) is correct because after simplifying the given equation and solving for cos⁡2x\cos 2x, we can find cos⁡4x\cos 4x which equals -7/9.
10 Medium
10 of 52
Application +4 | -1
The value of tan⁡  821∘2\tan \,\,82\frac{{1^\circ }}{2}, is
View answer and explanation
Correct option: A
2+3+4+6\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6
Option A (2+3+4+6\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6) is the correct answer because it is obtained by simplifying the expression for tan⁡82.5∘\tan 82.5^\circ using trigonometric identities.
11 Medium
11 of 52
Application +4 | -1
If sin⁡α=336625\sin \alpha=\frac{336}{625} and 450∘<α<540∘450^{\circ}<\alpha<540^{\circ}, then sin⁡(α/4)\sin (\alpha / 4) is equal to
View answer and explanation
Correct option: A
152\frac{1}{5 \sqrt{2}}
Option A (152\frac{{1}}{{5\sqrt{{2}}}}) is correct because by using the given information and applying trigonometric identities repeatedly, the value of sin⁡(α/4)\sin(\alpha/4) is found to be 152\frac{{1}}{{5\sqrt{{2}}}}.
12 Medium
12 of 52
Application +4 | -1
The value of (1+cos⁡  π6)  (1+cos⁡  π3)  (1+cos⁡  2π3)  (1+cos⁡  7π6)\left( {1 + \cos \,\,\frac{\pi }{6}} \right)\,\,\left( {1 + \cos \,\,\frac{\pi }{3}} \right)\,\,\left( {1 + \cos \,\,\frac{{2\pi }}{3}} \right)\,\,\left( {1 + \cos \,\,\frac{{7\pi }}{6}} \right) is
View answer and explanation
Correct option: A
316\frac{3}{{16}}
Option A (316\frac{3}{{16}}) is correct because it is derived by substituting the values of the cosine function and simplifying the expression.
13 Medium
13 of 52
Application +4 | -1
If sin⁡α=336625\sin \alpha=\frac{336}{625} and 450∘<α<540∘450^{\circ} < \alpha < 540^{\circ}, then sin⁡(α/4)\sin (\alpha / 4) is equal to
View answer and explanation
Correct option: A
152\frac{1}{5 \sqrt{2}}
Option A (152\frac{1}{5 \sqrt{2}}) is correct because by using the given information and trigonometric identities, we can find the value of sin⁡(α/4)\sin(\alpha / 4).
14 Medium
14 of 52
Analyzing +4 | -1
If θ=π/7\theta=\pi / 7, then cos⁡θ+cos⁡3θ+cos⁡5θ\cos \theta+\cos 3 \theta+\cos 5 \theta is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the sum of the cosines can be simplified to 12\frac{1}{2} when θ=π7\theta = \frac{\pi}{7}.
15 Medium
15 of 52
Application +4 | -1
If tan⁡  x=ba\tan \,\,x = \frac{b}{a}, then the value of a  cos⁡  2x+  b  sin⁡  2xa\,\,\cos \,\,2x + \,\,b\,\,\sin \,\,2x is
View answer and explanation
Correct option: D
aa
Option D is correct because by substituting the value of tan x and simplifying, the expression reduces to 'a'.

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16 Medium
16 of 52
Analyzing +4 | -1
The value of cos⁡π11cos⁡2π11cos⁡3π11………cos⁡11π11\cos \frac{\pi}{11} \cos \frac{2 \pi}{11} \cos \frac{3 \pi}{11} \ldots \ldots \ldots \cos \frac{11 \pi}{11} is equal to
View answer and explanation
Correct option: C
11024\frac{1}{{1024}}
Option C (11024\frac{1}{{1024}}) is correct because the product can be simplified using trigonometric identities and the sine product formula.
17 Medium
17 of 52
Analyzing +4 | -1
The expression cos⁡2(A−B)+cos⁡2B−2  cos⁡(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: C
dependent on AA
Option C (dependent on AA) is correct because the expression simplifies to sin⁡2A{\sin ^2}A, which depends only on AA.
18 Medium
18 of 52
Application +4 | -1
θ\theta and ϕ\phi are acute angles. If sin⁡θ=12\sin \theta=\frac{1}{2} and cos⁡ϕ=13\cos \phi=\frac{1}{3}, then θ+ϕ∈…\theta+\phi \in \ldots.
View answer and explanation
Correct option: B
(π2,2π3)\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right)
Option B ((π2,2π3)\left(\frac{{\pi}}{{2}}, \frac{{2 \pi}}{{3}}\right)) is correct because given the values of sin⁡θ\sin \theta and cos⁡ϕ\cos \phi, the sum θ+ϕ\theta + \phi falls within this interval.
19 Medium
19 of 52
Application +4 | -1
If 0<A<π60 < A < \frac{\pi }{6} and sin⁡A+cos⁡A=72\sin A + \cos A = \frac{{\sqrt 7 }}{2}, then tan⁡A2=\tan \frac{A}{2} =
View answer and explanation
Correct option: A
7−23\frac{{\sqrt 7 - 2}}{3}
Option A (7−23\frac{{\sqrt 7 - 2}}{3}) is correct because it is the solution for tan⁡A2\tan \frac{A}{2} derived from the given trigonometric equation and the constraint on A.
20 Medium
20 of 52
Application +4 | -1
If 0<θ<π0 < \theta < \pi, then 2+  2  +2  +…+2+2  cos⁡  θ\sqrt {2 + \,\,\sqrt {2\,\, + \sqrt {2\,\, + \ldots + \sqrt {2 + 2\,\,\cos \,\,\theta } } } } there being n number of 2′ s\mathrm{2'}\,\mathrm{s} is equal to
View answer and explanation
Correct option: A
2  cos⁡  θ2n2\,\,\cos \,\,\frac{\theta }{{{2^n}}}
Option A (2cos⁡θ2n2\cos \frac{\theta}{{{{2^n}}}}) is correct because it represents the simplified form of the nested square root expression after applying the identity 1+cos⁡θ=2cos⁡2θ21 + \cos \theta = 2\cos^2 \frac{\theta}{2} n times.
21 Medium
21 of 52
Application +4 | -1
The value of sec⁡8θ−1sec⁡4θ−1\frac{\sec 8 \theta-1}{\sec 4 \theta-1} is
View answer and explanation
Correct option: C
tan⁡8θtan⁡2θ\frac{\tan 8 \theta}{\tan 2 \theta}
Option C (tan⁡8θtan⁡2θ\frac{{\tan 8 \theta}}{{\tan 2 \theta}}) is correct because the expression simplifies to tan⁡8θtan⁡2θ\frac{{\tan 8 \theta}}{{\tan 2 \theta}} using trigonometric identities.
22 Medium
22 of 52
Application +4 | -1
If k=sin⁡π/18sin⁡5π′18sin⁡7π/18k=\sin \pi / 18 \sin 5 \pi^{\prime} 18 \sin 7 \pi / 18, then the numerical value of kk is equal to
View answer and explanation
Correct option: C
1/81 / 8
Option C (1/81 / 8) is correct because it is obtained by using the product-to-sum trigonometric identities to simplify the expression.
23 Medium
23 of 52
Application +4 | -1
α,  β(α≠β)\alpha ,\;\beta \left( {\alpha \ne \beta } \right) satisfy the equation a  cos⁡θ+b  sin⁡θ=ca\,\,\cos \theta + b\,\,\sin \theta = c, then the value of tan⁡(α+β2)\tan \left( {\frac{{\alpha + \beta }}{2}} \right), is

View answer and explanation
Correct option: A
b/ab/a
Option A (b/ab/a) is correct because it represents the value of tan⁡(α+β2)\tan \left( {\frac{{\alpha + \beta }}{2}} \right) derived from transforming the given equation and using trigonometric identities.
24 Medium
24 of 52
Application +4 | -1
If a2−2acos⁡x+1=674a^2-2 a \cos x+1=674 and tan⁡(x/2)=7\tan (x / 2)=7 then the integral value of aa is
View answer and explanation
Correct option: A
25
Option A (25) is correct because solving the equation a2−2acos⁡x+1=674a^2-2 a \cos x+1=674 with tan⁡(x/2)=7\tan (x / 2)=7 yields a=25a=25.
25 Medium
25 of 52
Application +4 | -1
If α,  β\mathrm{\alpha },\mathrm{\;\beta } are different values of xx satisfying acos⁡x+bsin⁡x=ca\cos x + b\sin x = c, then tan⁡(  α+β2)\tan \left( {\;\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right) is equal to

View answer and explanation
Correct option: C
ba\frac{b}{a}
Option C (ba\frac{b}{a}) is correct because it is derived by using the tangent half-angle substitution and solving the resulting quadratic equation.
26 Medium
26 of 52
Application +4 | -1
If cos⁡  20∘=k\cos \,\,20^\circ = k and cos   x=2k2−1cos\;\,x = 2{k^2} - 1, then the possible values of xx between 0∘0^\circ and 360∘360^\circ are
View answer and explanation
Correct option: C
40∘40^\circ and 320∘320^\circ
Option C (40∘40^\circ and 320∘320^\circ) is correct because using the double angle formula for cosine, we find that cos(x) = cos(40), leading to x = 40 and x = 360 - 40 = 320.
27 Medium
27 of 52
Application +4 | -1
If sin⁡θ1−sin⁡θ2=a\sin \theta_1-\sin \theta_2=a and cos⁡θ1+cos⁡θ2=b\cos \theta_1+\cos \theta_2=b, then
View answer and explanation
Correct option: B
a2+b2≤4\mathrm{a}^2+\mathrm{b}^2 \leq 4
Option B (a2+b2≤4a^2+b^2 \leq 4) is correct because by squaring and adding the given equations, we can simplify the expression to a2+b2=2+2cos⁡(θ1+θ2)a^2 + b^2 = 2 + 2 \cos(\theta_1 + \theta_2), which is always less than or equal to 4.
28 Medium
28 of 52
Analyzing +4 | -1
If x=y  cos⁡  2 π3=z  cos⁡4 π3x = y\,\,\cos \,\,\frac{{2\,\pi }}{3} = z\,\,\cos \frac{{4\,\pi }}{3}, then xy+yz+zx=xy + yz + zx =
View answer and explanation
Correct option: B
0
Option B (0) is correct because when x, y, and z are expressed in terms of a common variable, the expression xy + yz + zx simplifies to 0.
29 Medium
29 of 52
Analyzing +4 | -1
The value of 36(4cos⁡29∘−1)(4cos⁡227∘−1)36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right) (4cos⁡281∘−1)(4cos⁡2243∘−1)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right) is
View answer and explanation
Correct option: D
36
Option D, 36, is the correct value of the given expression. This is found by using trigonometric identities to simplify the expression.
30 Medium
30 of 52
Application +4 | -1
If a2+12a=cos⁡  θ\frac{{{a^2} + 1}}{{2a}} = \cos \,\,\theta, then a6+12a3=\frac{{{a^6} + 1}}{{2{a^3}}} =
View answer and explanation
Correct option: D
cos⁡  3θ\cos \,\,3\theta
Option D (cos⁡ 3θ\cos \,3\theta) is correct because by manipulating the given equation and using trigonometric identities, we can find that a6+12a3=cos⁡ 3θ\frac{{a^6 + 1}}{{2{a^3}}} = \cos \,3\theta.
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