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Trigonometric Ratio and Identities

JEE Mathematics · 51 questions · Page 2 of 6 · Click an option or "Show Solution" to reveal answer

Q11
Let the range of the function f(x)=6+16cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6x,x∈Rf(x)=6+16 \cos x \cdot \cos \left(\dfrac{\pi}{3}-x\right) \cdot \cos \left(\dfrac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x, x \in \mathbf{R} be [α,β][\alpha, \beta]. Then the distance of the point (α,β)(\alpha, \beta) from the line 3x+4y+12=03 x+4 y+12=0 is :
A 11
B 10
C 8
D 9
Correct Answer
Option A
Solution
f(x)=6+16(14cos⁡3x)sin⁡3x⋅cos⁡6x=6+4cos⁡3xsin⁡3xcos⁡6x=6+sin⁡12x Range of f(x) is [5, 7] (α,β)≡(5,7) distance =∣15+28+125∣=11\begin{aligned} &\begin{aligned} f(x) & =6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x \\ & =6+4 \cos 3 x \sin 3 x \cos 6 x \\ & =6+\sin 12 x \end{aligned}\\ &\text{ Range of } f(x) \text{ is [5, 7] }\\ &\begin{aligned} & (\alpha, \beta) \equiv(5,7) \\ & \text{ distance }=\left|\frac{15+28+12}{5}\right|=11 \end{aligned} \end{aligned}
Q12
The value of cos⁡π22.cos⁡π23 .....cos⁡π210.sin⁡π210\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}} is -
A 1256{1 \over {256}}
B 12{1 \over {2}}
C 11024{1 \over {1024}}
D 1512{1 \over {512}}
Correct Answer
Option D
Solution

Given

cos⁡π22.cos⁡π23 .....cos⁡π210.sin⁡π210\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}

Let

π210 = θ{\pi \over {{2^{10}}}}\, = \,\theta

∴\therefore

π29 = 2θ{\pi \over {{2^9}}}\, = \,2\theta
π28 = 22θ{\pi \over {{2^8}}}\, = \,{2^2}\theta
π27 = 23θ{\pi \over {{2^7}}}\, = \,{2^3}\theta

. .

π22 = 28θ{\pi \over {{2^2}}}\, = \,{2^8}\theta

So given term becomes,

cos⁡28θ.cos⁡27θ.....cos⁡θ\cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta
.sin⁡π210.\sin {\pi \over {{2^{10}}}}

=

(cos⁡θ.cos⁡2θ......cos⁡28θ)sin⁡π210(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}

=

sin⁡29θ29sin⁡θ.sin⁡π210{{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}

=

sin⁡29(π210)29sin⁡π210.sin⁡π210{{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}

=

sin⁡(π2)29{{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}

=

129{1 \over {{2^9}}}

=

1512{1 \over {512}}

Note :

(cos⁡θ.cos⁡2θ......cos⁡2n−1θ)(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )

=

sin⁡2nθ2nsin⁡θ{{\sin {2^n}\theta } \over {{2^n}\sin \theta }}
Q13
2sin⁡(π22)sin⁡(3π22)sin⁡(5π22)sin⁡(7π22)sin⁡(9π22)2 \sin \left(\dfrac{\pi}{22}\right) \sin \left(\dfrac{3 \pi}{22}\right) \sin \left(\dfrac{5 \pi}{22}\right) \sin \left(\dfrac{7 \pi}{22}\right) \sin \left(\dfrac{9 \pi}{22}\right) is equal to :
A 316\dfrac{3}{16}
B 116\dfrac{1}{16}
C 132\dfrac{1}{32}
D 932\dfrac{9}{32}
Correct Answer
Option B
Solution
2sin⁡π22sin⁡3π22sin⁡5π22sin⁡7π22sin⁡9π222\sin {\pi \over {22}}\sin {{3\pi } \over {22}}\sin {{5\pi } \over {22}}\sin {{7\pi } \over {22}}\sin {{9\pi } \over {22}}
=2sin⁡(11π−10π22)sin⁡(11π−8π22)sin⁡(11π−6π22)sin⁡(11π−4π22)sin⁡(11π−2π22)= 2\sin \left( {{{11\pi - 10\pi } \over {22}}} \right)\sin \left( {{{11\pi - 8\pi } \over {22}}} \right)\sin \left( {{{11\pi - 6\pi } \over {22}}} \right)\sin \left( {{{11\pi - 4\pi } \over {22}}} \right)\sin \left( {{{11\pi - 2\pi } \over {22}}} \right)
=2cos⁡π11cos⁡2π11cos⁡3π11cos⁡4π11cos⁡5π11= 2\cos {\pi \over {11}}\cos {{2\pi } \over {11}}\cos {{3\pi } \over {11}}\cos {{4\pi } \over {11}}\cos {{5\pi } \over {11}}
=2sin⁡32π1125sin⁡π11= {{2\sin {{32\pi } \over {11}}} \over {{2^5}\sin {\pi \over {11}}}}
=116= {1 \over {16}}
Q14
If tan⁡15∘+1tan⁡75∘+1tan⁡105∘+tan⁡195∘=2a\tan 15^\circ + {1 \over {\tan 75^\circ }} + {1 \over {\tan 105^\circ }} + \tan 195^\circ = 2a, then the value of (a+1a)\left( {a + {1 \over a}} \right) is :
A 5−3235 - {3 \over 2}\sqrt 3
B 4−234 - 2\sqrt 3
C 2
D 4
Correct Answer
Option D
Solution
tan⁡15∘+tan⁡15∘−tan⁡15∘+tan⁡15∘\tan 15^\circ + \tan 15^\circ - \tan 15^\circ + \tan 15^\circ
=2tan⁡15∘= 2\tan 15^\circ
=2(2−3)=2a⇒a=2−3= 2\left( {2 - \sqrt 3 } \right) = 2a \Rightarrow a = 2 - \sqrt 3

∴\therefore

1a+a⇒(2+3)+(2−3)=4{1 \over a} + a \Rightarrow \left( {2 + \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) = 4
Q15
The set of all values of λ\lambda for which the equation cos⁡22x−2sin⁡4x−2cos⁡2x=λ{\cos ^2}2x - 2{\sin ^4}x - 2{\cos ^2}x = \lambda has a real solution xx, is :
A [−2,−1]\left[ { - 2, - 1} \right]
B [−32,−1]\left[ { - {3 \over 2}, - 1} \right]
C [−2,−32]\left[ { - 2, - {3 \over 2}} \right]
D [−1,−12]\left[ { - 1, - {1 \over 2}} \right]
Correct Answer
Option B
Solution

The given equation is

cos⁡22x−2sin⁡4x−2cos⁡2x=λ\cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x = \lambda

Using the trigonometric identities

cos⁡2x=1−sin⁡2x\cos^2x = 1 - \sin^2x

and

cos⁡22x=1−2sin⁡2x\cos^22x = 1 - 2\sin^2x

, we can rewrite the equation in terms of

cos⁡2x\cos^2x

:

λ=(2cos⁡2x−1)2−2(1−cos⁡2x)2−2cos⁡2x\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x

Simplify this equation :

λ=4cos⁡4x−4cos⁡2x+1−2(1−2cos⁡2x+cos⁡4x)−2cos⁡2x\lambda = 4 \cos ^4 x - 4 \cos ^2 x + 1 - 2(1 - 2 \cos ^2 x + \cos ^4 x) - 2 \cos ^2 x
λ=2cos⁡4x−2cos⁡2x−1\lambda = 2 \cos ^4 x - 2 \cos ^2 x - 1

We can factor out a 2 and rewrite this as :

λ=2(cos⁡4x−cos⁡2x−12)\lambda = 2(\cos ^4 x - \cos ^2 x - \frac{1}{2})
λ=2[(cos⁡2x−12)2−34]\lambda = 2[(\cos ^2 x - \frac{1}{2})^2 - \frac{3}{4}]

So λmax⁡=2[14−34]=2×(−24)=−1\lambda_{\max }=2\left[\dfrac{1}{4}-\dfrac{3}{4}\right]=2 \times\left(\dfrac{-2}{4}\right)=-1 (maximum value) and λmin⁡=2[0−34]=−32(\lambda_{\min }=2\left[0-\dfrac{3}{4}\right]=-\dfrac{3}{2}( minimum value )) So range of the value of xx is [−32,−1]\left[\dfrac{-3}{2},-1\right].

Q16
If the equation cos4 θ\theta + sin4 θ\theta + λ\lambda = 0 has real solutions for θ\theta , then λ\lambda lies in the interval :
A [−32,−54]\left[ { - {3 \over 2}, - {5 \over 4}} \right]
B (−12,−14]\left( { - {1 \over 2}, - {1 \over 4}} \right]
C (−54,−1]\left( { - {5 \over 4}, - 1} \right]
D [−1,−12]\left[ { - 1, - {1 \over 2}} \right]
Correct Answer
Option D
Solution

cos4 θ\theta + sin4 θ\theta + λ\lambda = 0 ⇒\Rightarrow 1 – 2sin2 θ\theta cos2 θ\theta = -λ\lambda ⇒\Rightarrow 1 -

12×4\frac{1}{2} \times 4

sin2 θ\theta cos2 θ\theta = -λ\lambda ⇒\Rightarrow 1 -

sin⁡22θ2\frac{\sin^{2} 2\theta }{2}

= -λ\lambda ⇒\Rightarrow2(λ\lambda + 1) = sin2 2θ\theta 0 ≤\le 2 (λ\lambda + 1) ≤\le 1 0 ≤\le (λ\lambda + 1) ≤\le

12\frac{1}{2}

-1 ≤\le λ\lambda ≤\le -

12\frac{1}{2}
Q17
If the value of 3cos⁡36∘+5sin⁡18∘5cos⁡36∘−3sin⁡18∘\dfrac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}} is a5−bc\dfrac{a \sqrt{5}-b}{c}, where a,b,ca, b, c are natural numbers and gcd⁡(a,c)=1\operatorname{gcd}(a, c)=1, then a+b+ca+b+c is equal to :
A 54
B 52
C 50
D 40
Correct Answer
Option B
Solution

To find the value of

3cos⁡36∘+5sin⁡18∘5cos⁡36∘−3sin⁡18∘\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}

in the form

a5−bc\frac{a \sqrt{5}-b}{c}

, we need to simplify the given expression.

Let's start by using some fundamental trigonometric identities.

We know that:

cos⁡36∘=5+14\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}
sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}

First, substitute these values into the expression:

3⋅5+14+5⋅5−145⋅5+14−3⋅5−14\frac{3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4}}{5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4}}

Simplify the numerator and the denominator: Numerator:

3⋅5+14+5⋅5−14=3(5+1)+5(5−1)4=35+3+55−54=85−24=25−123 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4} = \frac{3(\sqrt{5} + 1) + 5(\sqrt{5} - 1)}{4} = \frac{3\sqrt{5} + 3 + 5\sqrt{5} - 5}{4} = \frac{8\sqrt{5} - 2}{4} = 2 \sqrt{5} - \frac{1}{2}

Denominator:

5⋅5+14−3⋅5−14=5(5+1)−3(5−1)4=55+5−35+34=25+84=5+425 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4} = \frac{5(\sqrt{5} + 1) - 3(\sqrt{5} - 1)}{4} = \frac{5\sqrt{5} + 5 - 3\sqrt{5} + 3}{4} = \frac{2\sqrt{5} + 8}{4} = \frac{\sqrt{5}+4}{2}

Now combine the simplified numerator and denominator:

25−125+42=(25−12)⋅25+4=45−15+4\frac{2 \sqrt{5} - \frac{1}{2}}{\frac{\sqrt{5}+4}{2}} = \frac{(2 \sqrt{5} - \frac{1}{2}) \cdot 2}{\sqrt{5}+4} = \frac{4 \sqrt{5} - 1}{\sqrt{5}+4}

Rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:

(45−1)(5−4)(5+4)(5−4)\frac{(4 \sqrt{5} - 1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)}

The denominator simplifies to:

52−42=5−16=−11\sqrt{5}^2 - 4^2 = 5 - 16 = -11

The numerator simplifies to:

(45−1)(5−4)=(45⋅5−4⋅45−1⋅5+1⋅4)=(20−165−5+4)=24−175(4 \sqrt{5} - 1)(\sqrt{5}-4) = (4 \sqrt{5} \cdot \sqrt{5} - 4 \cdot 4 \sqrt{5} - 1 \cdot \sqrt{5} + 1 \cdot 4) = (20 - 16 \sqrt{5} - \sqrt{5} + 4) = 24 - 17 \sqrt{5}

Combining them, we get:

24−175−11=175−2411\frac{24 - 17\sqrt{5}}{-11} = \frac{17\sqrt{5}-24}{11}

Thus,

a=17,b=24,c=11a = 17, b = 24, c = 11

. Therefore,

a+b+c=17+24+11=52a + b + c = 17 + 24 + 11 = 52

. So the correct option is: Option B: 52

Q18
If sin⁡x=−35\sin x=-\dfrac{3}{5}, where $$\pi
A 109
B 108
C 19
D 18
Correct Answer
Option A
Solution

$$\begin{aligned} & \sin x=-\frac{3}{5} \text { where } \pi

Q19
Suppose θ∈[0,π4]\theta \in\left[0, \dfrac{\pi}{4}\right] is a solution of 4cos⁡θ−3sin⁡θ=14 \cos \theta-3 \sin \theta=1. Then cos⁡θ\cos \theta is equal to :
A 6−6(36−2)\dfrac{6-\sqrt{6}}{(3 \sqrt{6}-2)}
B 4(36+2)\dfrac{4}{(3 \sqrt{6}+2)}
C 6+6(36+2)\dfrac{6+\sqrt{6}}{(3 \sqrt{6}+2)}
D 4(36−2)\dfrac{4}{(3 \sqrt{6}-2)}
Correct Answer
Option D
Solution
4cos⁡θ−3sin⁡θ=14cos⁡θ−1=3sin⁡θ16cos⁡2θ+1−8cos⁡θ=9(1−cos⁡2θ)⇒25cos⁡2θ−8cos⁡θ−8=0⇒cos⁡θ=8±64+4×25×82.25=8±44+502.25\begin{aligned} & 4 \cos \theta-3 \sin \theta=1 \\ & 4 \cos \theta-1=3 \sin \theta \\ & 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\ & \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\ & \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\ & =\frac{8 \pm 4 \sqrt{4+50}}{2.25} \end{aligned}
=4±25425 As θ∈[0,π4]⇒cos⁡θ=4+6625=436−2\begin{aligned} & =\frac{4 \pm 2 \sqrt{54}}{25} \\ & \text{ As } \theta \in\left[0, \frac{\pi}{4}\right] \\ & \Rightarrow \cos \theta=\frac{4+6 \sqrt{6}}{25}=\frac{4}{3 \sqrt{6}-2} \end{aligned}
Q20
If sin⁡x+sin⁡2x=1\sin x + \sin^2 x = 1, x∈(0,π2)x \in \left(0, \dfrac{\pi}{2}\right), then (cos⁡12x+tan⁡12x)+3(cos⁡10x+tan⁡10x+cos⁡8x+tan⁡8x)+(cos⁡6x+tan⁡6x)(\cos^{12} x + \tan^{12} x) + 3(\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x) is equal to:
A 3
B 4
C 2
D 1
Correct Answer
Option C
Solution
sin⁡x+sin⁡2x=1⇒sin⁡x=cos⁡2x⇒tan⁡x=cos⁡x∴ Given expression =2cos⁡12x+6[cos⁡10x+cos⁡8x]+2cos⁡6x=2[sin⁡6x+3sin⁡5x+3sin⁡4x+sin⁡3x]=2sin⁡3x[(sin⁡x+1)3]=2[sin⁡2x+sin⁡x]3=2\begin{aligned} &\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}\\ &\therefore \text{ Given expression }\\ &\begin{aligned} & =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \\ & =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \\ & =2 \sin ^3 x\left[(\sin x+1)^3\right] \\ & =2\left[\sin ^2 x+\sin x\right]^3 \\ & =2 \end{aligned} \end{aligned}
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