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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 11–20 (part 2 of 8)

  1. Exercise 11

    If tan⁡(A+B)=p,tan⁡( A−B)=q\displaystyle \tan (\mathrm{A}+\mathrm{B})=p, \tan (\mathrm{~A}-\mathrm{B})=q, then show that tan⁡2 A=p+q1−pq\displaystyle \tan 2 \mathrm{~A}=\frac{p+q}{1-p q} [Hint: Use 2 A=(A+B)+(A−B)\displaystyle 2 \mathrm{~A}=(\mathrm{A}+\mathrm{B})+(\mathrm{A}-\mathrm{B}) ]

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    Write \(\displaystyle 2A=(A+B)+(A-B)\).\[\tan 2A=\tan\big[(A+B)+(A-B)\big] \] \[\tan 2A=\frac{\tan(A+B)+\tan(A-B)}{1-\tan(A+B)\tan(A-B)} \] \[\tan 2A=\frac{p+q}{1-pq} \]Answer: \(\displaystyle \tan 2A=\dfrac{p+q}{1-pq}\)

  2. Exercise 12

    If cos⁡α+cos⁡β=0=sin⁡α+sin⁡β\displaystyle \cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta, then prove that cos⁡2α+cos⁡2β=−2cos⁡(α+β)\displaystyle \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta). [Hint: (cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2=0\displaystyle (\cos \alpha+\cos \beta)^2-(\sin \alpha+\sin \beta)^2=0]

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    Both sums are \(\displaystyle 0\), so the difference of their squares is \(\displaystyle 0\).\[(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2=0 \] \[(\cos^2\alpha-\sin^2\alpha)+(\cos^2\beta-\sin^2\beta)+2(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=0 \] \[\cos 2\alpha+\cos 2\beta+2\cos(\alpha+\beta)=0 \] \[\cos 2\alpha+\cos 2\beta=-2\cos(\alpha+\beta) \]Answer: \(\displaystyle \cos 2\alpha+\cos 2\beta=-2\cos(\alpha+\beta)\)

  3. Exercise 13

    If sin⁡(x+y)sin⁡(x−y)=a+ba−b\displaystyle \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}, then show that tan⁡xtan⁡y=ab\displaystyle \frac{\tan x}{\tan y}=\frac{a}{b} [Hint: Use Componendo and Dividendo].

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    Apply componendo and dividendo: if \(\displaystyle \frac pq=\frac rs\), then \(\displaystyle \frac{p+q}{p-q}=\frac{r+s}{r-s}\).\[\frac{\sin(x+y)+\sin(x-y)}{\sin(x+y)-\sin(x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)} \] \[\sin(x+y)+\sin(x-y)=2\sin x\cos y \] \[\sin(x+y)-\sin(x-y)=2\cos x\sin y \] \[\frac{2\sin x\cos y}{2\cos x\sin y}=\frac{2a}{2b} \] \[\frac{\tan x}{\tan y}=\frac ab \]Answer: \(\displaystyle \dfrac{\tan x}{\tan y}=\dfrac ab\)

  4. Exercise 14

    If tan⁡θ=sin⁡α−cos⁡αsin⁡α+cos⁡α\displaystyle \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, then show that sin⁡α+cos⁡α=2cos⁡θ\displaystyle \sin \alpha+\cos \alpha=\sqrt{2} \cos \theta. [Hint: Express tan⁡θ=tan⁡(α−π4)θ=α−π4\displaystyle \tan \theta=\tan \left(\alpha-\frac{\pi}{4}\right) \quad \theta=\alpha-\frac{\pi}{4} ]

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    Divide numerator and denominator by \(\displaystyle \cos\alpha\).\[\tan\theta=\frac{\tan\alpha-1}{\tan\alpha+1}=\frac{\tan\alpha-\tan\frac\pi4}{1+\tan\alpha\tan\frac\pi4} \] \[\tan\theta=\tan\left(\alpha-\frac\pi4\right)\ \Rightarrow\ \theta=\alpha-\frac\pi4 \] \[\sin\alpha+\cos\alpha=\sqrt2\left(\sin\alpha\cos\frac\pi4+\cos\alpha\sin\frac\pi4\right)=\sqrt2\sin\left(\alpha+\frac\pi4\right) \] \[\sqrt2\sin\left(\alpha+\frac\pi4\right)=\sqrt2\cos\left(\frac\pi4-\alpha\right)=\sqrt2\cos\left(\alpha-\frac\pi4\right)=\sqrt2\cos\theta \](\(\displaystyle \theta+\pi\) has the same tangent but the opposite cosine, so \(\displaystyle \theta=\alpha-\frac\pi4\) is the angle meant.)Answer: \(\displaystyle \sin\alpha+\cos\alpha=\sqrt2\cos\theta\)

  5. Exercise 15

    If sin⁡θ+cos⁡θ=1\displaystyle \sin \theta+\cos \theta=1, then find the general value of θ\displaystyle \theta.

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    NCERT’s answer
    \(\displaystyle \theta=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4}\)
    Combine into a single cosine.\[\sin\theta+\cos\theta=\sqrt2\cos\left(\theta-\frac\pi4\right)=1 \] \[\cos\left(\theta-\frac\pi4\right)=\frac1{\sqrt2}=\cos\frac\pi4 \] \[\theta-\frac\pi4=2n\pi\pm\frac\pi4 \] \[\theta=2n\pi+\frac\pi2\quad\text{or}\quad\theta=2n\pi \]Answer: \(\displaystyle \theta=2n\pi\) or \(\displaystyle \theta=2n\pi+\dfrac\pi2\), \(\displaystyle n\in\mathbb Z\)

  6. Exercise 16

    Find the most general value of θ\displaystyle \theta satisfying the equation tan⁡θ=−1\displaystyle \tan \theta=-1 and cos⁡θ=12\displaystyle \cos \theta=\frac{1}{\sqrt{2}}.

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    NCERT’s answer
    \(\displaystyle \theta=2 n \pi+\frac{7 \pi}{4}\)
    \(\displaystyle \tan\theta<0\) puts \(\displaystyle \theta\) in the second or fourth quadrant; \(\displaystyle \cos\theta>0\) selects the fourth.\[\tan\theta=-1\ \Rightarrow\ \theta=m\pi-\frac\pi4 \] \[\cos\theta=\frac1{\sqrt2}\ \Rightarrow\ \theta=2n\pi\pm\frac\pi4 \] \[\text{common values: } \theta=\frac{7\pi}4 \] \[\theta=2n\pi+\frac{7\pi}4=2n\pi-\frac\pi4 \]Answer: \(\displaystyle \theta=2n\pi+\dfrac{7\pi}4\), \(\displaystyle n\in\mathbb Z\)

  7. Exercise 17

    If cot⁡θ+tan⁡θ=2cosec⁡θ\displaystyle \cot \theta+\tan \theta=2 \operatorname{cosec} \theta, then find the general value of θ\displaystyle \theta.

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    NCERT’s answer
    \(\displaystyle \theta=2 n \pi \pm \frac{\pi}{3}\)
    \[\cot\theta+\tan\theta=\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta} \] \[\frac{1}{\sin\theta\cos\theta}=\frac{2}{\sin\theta} \] Cancel \(\displaystyle \sin\theta\ne0\) (\(\displaystyle \operatorname{cosec}\theta\) is defined): \[\cos\theta=\frac12=\cos\frac{\pi}{3} \] \[\theta=2n\pi\pm\frac{\pi}{3},\quad n\in\mathbb{Z} \] Answer: \(\displaystyle \theta=2n\pi\pm\dfrac{\pi}{3}\), \(\displaystyle n\in\mathbb{Z}\).

  8. Exercise 18

    If 2sin⁡2θ=3cos⁡θ\displaystyle 2 \sin ^2 \theta=3 \cos \theta, where 0≤θ≤2π\displaystyle 0 \leq \theta \leq 2 \pi, then find the value of θ\displaystyle \theta.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \theta=\frac{\pi}{3}, \frac{5 \pi}{3}\)
    \[2\sin^2\theta=3\cos\theta \] \[2(1-\cos^2\theta)=3\cos\theta \] \[2\cos^2\theta+3\cos\theta-2=0 \] \[(2\cos\theta-1)(\cos\theta+2)=0 \] \(\displaystyle \cos\theta=-2\) is impossible, so \[\cos\theta=\frac12 \] \[\theta=\frac{\pi}{3},\ \frac{5\pi}{3}\quad (0\le\theta\le2\pi) \] Answer: \(\displaystyle \theta=\dfrac{\pi}{3}\) or \(\displaystyle \dfrac{5\pi}{3}\).

  9. Exercise 19

    If sec⁡xcos⁡5x+1=0\displaystyle \sec x \cos 5 x+1=0, where 0<x≤π2\displaystyle 0<x \leq \frac{\pi}{2}, then find the value of x\displaystyle x.

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    \[\sec x\cos5x=-1 \;\Rightarrow\; \cos5x=-\cos x \quad(\cos x\ne0) \] \[\cos5x+\cos x=0 \] \[2\cos3x\cos2x=0 \] \[\cos3x=0:\quad 3x=\frac{\pi}{2},\frac{3\pi}{2}\;\Rightarrow\; x=\frac{\pi}{6},\frac{\pi}{2}\qquad(0<3x\le\tfrac{3\pi}{2}) \] \[\cos2x=0:\quad 2x=\frac{\pi}{2}\;\Rightarrow\; x=\frac{\pi}{4}\qquad(0<2x\le\pi) \] \(\displaystyle x=\dfrac{\pi}{2}\) is rejected because \(\displaystyle \sec x\) is undefined there. \[x=\frac{\pi}{6}:\ \sec x\cos5x=\frac{2}{\sqrt3}\cdot\left(-\frac{\sqrt3}{2}\right)=-1 \] \[x=\frac{\pi}{4}:\ \sec x\cos5x=\sqrt2\cdot\left(-\frac{1}{\sqrt2}\right)=-1 \] Answer: \(\displaystyle x=\dfrac{\pi}{6}\) or \(\displaystyle \dfrac{\pi}{4}\).

  10. Exercise 20

    If sin⁡(θ+α)=a\displaystyle \sin (\theta+\alpha)=a and sin⁡(θ+β)=b\displaystyle \sin (\theta+\beta)=b, then prove that cos⁡2(α−β)−4abcos⁡(α−β)=\displaystyle \cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta)= 1−2a2−2b2\displaystyle 1-2 a^2-2 b^2 [Hint: Express cos⁡(α−β)=cos⁡((θ+α)−(θ+β))\displaystyle \cos (\alpha-\beta)=\cos ((\theta+\alpha)-(\theta+\beta)) ]

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    Let \(\displaystyle A=\theta+\alpha\), \(\displaystyle B=\theta+\beta\); then \(\displaystyle \alpha-\beta=A-B\), \(\displaystyle \sin A=a\), \(\displaystyle \sin B=b\). \[\cos(\alpha-\beta)=\cos A\cos B+\sin A\sin B=\cos A\cos B+ab \] \[\cos(\alpha-\beta)-ab=\cos A\cos B \] \[\left[\cos(\alpha-\beta)-ab\right]^2=(1-a^2)(1-b^2) \] \[\cos^2(\alpha-\beta)-2ab\cos(\alpha-\beta)+a^2b^2=1-a^2-b^2+a^2b^2 \] \[\cos^2(\alpha-\beta)-2ab\cos(\alpha-\beta)=1-a^2-b^2 \] \[\cos2(\alpha-\beta)-4ab\cos(\alpha-\beta)=2\cos^2(\alpha-\beta)-1-4ab\cos(\alpha-\beta) \] \[=2\left[\cos^2(\alpha-\beta)-2ab\cos(\alpha-\beta)\right]-1 \] \[=2(1-a^2-b^2)-1=1-2a^2-2b^2 \]