Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Trigonometric Functions and Equations question

2012 · Shift 1 · Q21
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Trigonometric Functions and Equations
  5. /2012 · Shift 1 · Q21

Trigonometric Functions and Equations question

2012 · Shift 1 · Q21

JEE AdvancedMathematicsTrigonometric Functions and EquationsMultiple correct+4 / −1
Let θ, φ ∈ [0,2π]\theta ,\,\varphi \, \in \,\left[ {0,2\pi } \right]θ,φ∈[0,2π] be such that 2cos⁡θ(1−sin⁡ φ)=sin⁡2θ  (tan⁡θ2+cot⁡θ2)cos⁡φ−1, tan⁡(2π−θ)>02\cos \theta \left( {1 - \sin \,\varphi } \right) = {\sin ^2}\theta \,\,\left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)\cos \varphi - 1,\,\tan \left( {2\pi - \theta } \right) \gt 02cosθ(1−sinφ)=sin2θ(tan2θ​+cot2θ​)cosφ−1,tan(2π−θ)>0 and −1<sin⁡θ <−32,- 1 \lt \sin \theta \, \lt - {{\sqrt 3 } \over 2},−1<sinθ<−23​​, then φ\varphiφ cannot satisfy
  1. A
    0<φ<π20 \lt \varphi \lt {\pi \over 2}0<φ<2π​
  2. B
    π2<φ<4π3{\pi \over 2} \lt \varphi \lt {{4\pi } \over 3}2π​<φ<34π​
  3. C
    4π3<φ<3π2{{4\pi } \over 3} \lt \varphi \lt {{3\pi } \over 2}34π​<φ<23π​
  4. D
    3π2<φ<2π{{3\pi } \over 2} \lt \varphi \lt 2\pi23π​<φ<2π
View written solutionFree

Correct answer: C, D, A

  1. Simplify the given equation

Given 2cos⁡θ(1−sin⁡φ)=sin⁡2θ(tan⁡θ2+cot⁡θ2)cos⁡φ−1.2\cos\theta(1-\sin\varphi)=\sin^2\theta\left(\tan\frac\theta2+\cot\frac\theta2\right)\cos\varphi-1.2cosθ(1−sinφ)=sin2θ(tan2θ​+cot2θ​)cosφ−1.

Use the identity tan⁡x+cot⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x=2sin⁡2x.\tan x+\cot x=\frac{\sin^2 x+\cos^2 x}{\sin x\cos x}=\frac{1}{\sin x\cos x}=\frac{2}{\sin 2x}.tanx+cotx=sinxcosxsin2x+cos2x​=sinxcosx1​=sin2x2​.

So for x=θ2x=\frac\theta2x=2θ​, tan⁡θ2+cot⁡θ2=2sin⁡θ.\tan\frac\theta2+\cot\frac\theta2=\frac{2}{\sin\theta}.tan2θ​+cot2θ​=sinθ2​.

Hence sin⁡2θ(tan⁡θ2+cot⁡θ2)=sin⁡2θ⋅2sin⁡θ=2sin⁡θ.\sin^2\theta\left(\tan\frac\theta2+\cot\frac\theta2\right)=\sin^2\theta\cdot \frac{2}{\sin\theta}=2\sin\theta.sin2θ(tan2θ​+cot2θ​)=sin2θ⋅sinθ2​=2sinθ.

Therefore the equation becomes 2cos⁡θ(1−sin⁡φ)=2sin⁡θcos⁡φ−1.2\cos\theta(1-\sin\varphi)=2\sin\theta\cos\varphi-1.2cosθ(1−sinφ)=2sinθcosφ−1.

Rearrange: 2cos⁡θ−2cos⁡θsin⁡φ−2sin⁡θcos⁡φ=−1.2\cos\theta-2\cos\theta\sin\varphi-2\sin\theta\cos\varphi=-1.2cosθ−2cosθsinφ−2sinθcosφ=−1.

Using sin⁡φcos⁡θ+cos⁡φsin⁡θ=sin⁡(θ+φ),\sin\varphi\cos\theta+\cos\varphi\sin\theta=\sin(\theta+\varphi),sinφcosθ+cosφsinθ=sin(θ+φ), we get 2cos⁡θ−2sin⁡(θ+φ)=−1,2\cos\theta-2\sin(\theta+\varphi)=-1,2cosθ−2sin(θ+φ)=−1, so 2sin⁡(θ+φ)=2cos⁡θ+1,2\sin(\theta+\varphi)=2\cos\theta+1,2sin(θ+φ)=2cosθ+1, that is, sin⁡(θ+φ)=cos⁡θ+12.\boxed{\sin(\theta+\varphi)=\cos\theta+\frac12}. sin(θ+φ)=cosθ+21​​.

  1. Find the range of θ\thetaθ from the conditions

We are given −1<sin⁡θ<−32.-1<\sin\theta<-\frac{\sqrt3}{2}.−1<sinθ<−23​​.

So θ\thetaθ lies where sine is negative and less than −32-\frac{\sqrt3}{2}−23​​, i.e. θ∈(4π3,5π3).\theta\in\left(\frac{4\pi}{3},\frac{5\pi}{3}\right).θ∈(34π​,35π​).

Also, tan⁡(2π−θ)>0.\tan(2\pi-\theta)>0.tan(2π−θ)>0. Since tan⁡(2π−θ)=−tan⁡θ,\tan(2\pi-\theta)=-\tan\theta,tan(2π−θ)=−tanθ, this means tan⁡θ<0.\tan\theta<0.tanθ<0.

In (4π3,5π3)\left(\frac{4\pi}{3},\frac{5\pi}{3}\right)(34π​,35π​), tangent is:

  • positive in quadrant III,
  • negative in quadrant IV.

Hence θ∈(3π2,5π3).\boxed{\theta\in\left(\frac{3\pi}{2},\frac{5\pi}{3}\right)}.θ∈(23π​,35π​)​.

Therefore θ\thetaθ is in quadrant IV.

So cos⁡θ∈(12,0]?\cos\theta\in\left(\frac12,0\right]?cosθ∈(21​,0]? More precisely, for θ∈(3π2,5π3)\theta\in\left(\frac{3\pi}{2},\frac{5\pi}{3}\right)θ∈(23π​,35π​), cos⁡θ∈(0,12).\cos\theta\in\left(0,\frac12\right).cosθ∈(0,21​).

Thus cos⁡θ+12∈(12,1).\cos\theta+\frac12\in\left(\frac12,1\right).cosθ+21​∈(21​,1).

Therefore sin⁡(θ+φ)∈(12,1).\sin(\theta+\varphi)\in\left(\frac12,1\right).sin(θ+φ)∈(21​,1).

So θ+φ∈(π6,5π6)∪(13π6,17π6)(mod2π),\theta+\varphi\in\left(\frac\pi6,\frac{5\pi}6\right) \cup \left(\frac{13\pi}6,\frac{17\pi}6\right) \pmod{2\pi},θ+φ∈(6π​,65π​)∪(613π​,617π​)(mod2π), more usefully, θ+φ∈(π6,5π6) or (13π6,17π6).\theta+\varphi\in\left(\frac\pi6,\frac{5\pi}6\right) \text{ or } \left(\frac{13\pi}6,\frac{17\pi}6\right).θ+φ∈(6π​,65π​) or (613π​,617π​).

Since θ∈(3π2,5π3),\theta\in\left(\frac{3\pi}{2},\frac{5\pi}{3}\right),θ∈(23π​,35π​), subtracting θ\thetaθ gives possible φ\varphiφ ranges.

  1. Determine possible range of φ\varphiφ

Let α=θ+φ.\alpha=\theta+\varphi.α=θ+φ. Then sin⁡α∈(12,1),\sin\alpha\in\left(\frac12,1\right),sinα∈(21​,1), so for φ∈[0,2π]\varphi\in[0,2\pi]φ∈[0,2π], we examine:

Case 1:

α∈(π6,5π6).\alpha\in\left(\frac\pi6,\frac{5\pi}6\right).α∈(6π​,65π​). Then φ=α−θ.\varphi=\alpha-\theta.φ=α−θ. Since θ>3π2\theta>\frac{3\pi}{2}θ>23π​, this makes φ<0\varphi<0φ<0 in this case, impossible for φ∈[0,2π]\varphi\in[0,2\pi]φ∈[0,2π].

So this case gives no valid φ\varphiφ.

Case 2:

α∈(13π6,17π6).\alpha\in\left(\frac{13\pi}6,\frac{17\pi}6\right).α∈(613π​,617π​). Then φ=α−θ.\varphi=\alpha-\theta.φ=α−θ.

Now

=\left(\frac\pi2,\frac{4\pi}3\right).$$ But because the intervals are open and $\theta$ varies inside an open interval, all values strictly inside occur. Hence $$\boxed{\varphi\in\left(\frac\pi2,\frac{4\pi}3\right).}$$ --- 4. **Check the options** - **A:** $0<\varphi<\frac\pi2$ Not possible. - **B:** $\frac\pi2<\varphi<\frac{4\pi}3$ Possible. - **C:** $\frac{4\pi}3<\varphi<\frac{3\pi}2$ Not possible. - **D:** $\frac{3\pi}2<\varphi<2\pi$ Not possible. So $\varphi$ **cannot satisfy** A, C, D. Thus the correct options are $$\boxed{A,\ C,\ D}. $$ --- 5. **Compare with stored correct answer** Stored correct answer: C, D, A Derived answer: A, C, D These are the same set of options.
PreviousNext

More from Trigonometric Functions and Equations

  • The positive integer value of n>3 satisfying the equation sin(nπ​)1​=sin(n2π​)1​+sin(n3π​)1​ is2011 · Numerical
  • Let P={θ:sinθ−cosθ=2​cosθ} and Q={θ:sinθ+cosθ=2​sinθ} be two sets. Then2011 · MCQ
  • The number of values of θ in the interval, (−2π​,2π​) such that θe5nπ​ for n=0,±1,±2 and tanθ=cot5θ as well as sin2θ=cos4θ…2010 · Numerical
  • The number of all possible values of θ where 0<θ<π, for which the system of equations (y+z)cos3θ=(xyz)sin3θxsin3θ=y2cos3θ​+z2sin3θ​(xyz)sin3θ=(y+2z)cos3θ+ysin3θ…2010 · Numerical
  • The maximum value of the expression sin2θ+3sinθcosθ+5cos2θ1​ is2010 · Numerical
  • Two parallel chords of a circle of radius 2 are at a distance 3​+1 apart. If the chords subtend at the center , angles of kπ​ and k2π​, where k>0, then the value of [k] is [Note…2010 · Numerical
  • If 2sin4x​+3cos4x​=51​, then2009 · Multiple correct
  • For 0<θ<2π​, the solution (s) of m=1∑6​csc(θ+4(m−1)π​)csc(θ+4mπ​)=42​ is (are)2009 · Multiple correct