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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 21–30 (part 3 of 8)

  1. Exercise 21

    If cos⁡(θ+ϕ)=mcos⁡(θ−ϕ)\displaystyle \cos (\theta+\phi)=m \cos (\theta-\phi), then prove that tan⁡θ=1−m1+mcot⁡ϕ\displaystyle \tan \theta=\frac{1-m}{1+m} \cot \phi. [Hint: Express cos⁡(θ+ϕ)cos⁡(θ−ϕ)=m1\displaystyle \frac{\cos (\theta+\phi)}{\cos (\theta-\phi)}=\frac{m}{1} and apply Componendo and Dividendo]

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    \[\cos(\theta+\phi)=m\cos(\theta-\phi) \] Componendo and dividendo: \[\frac{\cos(\theta-\phi)-\cos(\theta+\phi)}{\cos(\theta-\phi)+\cos(\theta+\phi)}=\frac{\cos(\theta-\phi)-m\cos(\theta-\phi)}{\cos(\theta-\phi)+m\cos(\theta-\phi)}=\frac{1-m}{1+m} \] \[\cos(\theta-\phi)-\cos(\theta+\phi)=2\sin\theta\sin\phi,\qquad \cos(\theta-\phi)+\cos(\theta+\phi)=2\cos\theta\cos\phi \] \[\frac{2\sin\theta\sin\phi}{2\cos\theta\cos\phi}=\frac{1-m}{1+m} \] \[\tan\theta\tan\phi=\frac{1-m}{1+m} \] \[\tan\theta=\frac{1-m}{1+m}\cot\phi \]

  2. Exercise 22

    Find the value of the expression 3[sin⁡4(3π2−α)+sin⁡4(3π+α)]−2{sin⁡6(π2+α)+sin⁡6(5π−α)]\displaystyle 3\left[\sin ^4\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^4(3 \pi+\alpha)\right]-2\left\{\sin ^6\left(\frac{\pi}{2}+\alpha\right)+\sin ^6(5 \pi-\alpha)\right]

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    NCERT’s answer
    $\displaystyle 1$
    \[\sin\left(\tfrac{3\pi}{2}-\alpha\right)=-\cos\alpha,\qquad \sin(3\pi+\alpha)=-\sin\alpha \] \[\sin\left(\tfrac{\pi}{2}+\alpha\right)=\cos\alpha,\qquad \sin(5\pi-\alpha)=\sin\alpha \] \[E=3(\cos^4\alpha+\sin^4\alpha)-2(\cos^6\alpha+\sin^6\alpha) \] \[\sin^4\alpha+\cos^4\alpha=1-2\sin^2\alpha\cos^2\alpha \] \[\sin^6\alpha+\cos^6\alpha=1-3\sin^2\alpha\cos^2\alpha \] \[E=3-6\sin^2\alpha\cos^2\alpha-2+6\sin^2\alpha\cos^2\alpha=1 \] Answer: \(\displaystyle 1\).

  3. Exercise 23

    If acos⁡2θ+bsin⁡2θ=c\displaystyle a \cos 2 \theta+b \sin 2 \theta=c has α\displaystyle \alpha and β\displaystyle \beta as its roots, then prove that tan⁡α+tan⁡β=2ba+c\displaystyle \tan \alpha+\tan \beta=\frac{2 b}{a+c}. [Hint: Use the identities cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\displaystyle \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} and sin⁡2θ=2tan⁡θ1+tan⁡2θ\displaystyle \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} ].

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Put \(\displaystyle t=\tan\theta\): \[\cos2\theta=\frac{1-t^2}{1+t^2},\qquad \sin2\theta=\frac{2t}{1+t^2} \] \[a\,\frac{1-t^2}{1+t^2}+b\,\frac{2t}{1+t^2}=c \] \[a(1-t^2)+2bt=c(1+t^2) \] \[(a+c)t^2-2bt+(c-a)=0 \] \(\displaystyle \alpha,\beta\) satisfy the given equation, so \(\displaystyle \tan\alpha,\tan\beta\) are the two roots of this quadratic. \[\tan\alpha+\tan\beta=-\frac{-2b}{a+c}=\frac{2b}{a+c} \]

  4. Exercise 24

    If x=sec⁡ϕ−tan⁡ϕ\displaystyle x=\sec \phi-\tan \phi and y=cosec⁡ϕ+cot⁡ϕ\displaystyle y=\operatorname{cosec} \phi+\cot \phi then show that xy+x−y+1=0\displaystyle x y+x-y+1=0 [Hint: Find xy+1\displaystyle x y+1 and then show that x−y=−(xy+1)\displaystyle x-y=-(x y+1) ]

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[xy=\frac{1-\sin\phi}{\cos\phi}\cdot\frac{1+\cos\phi}{\sin\phi}=\frac{1+\cos\phi-\sin\phi-\sin\phi\cos\phi}{\sin\phi\cos\phi} \] \[xy+1=\frac{1+\cos\phi-\sin\phi}{\sin\phi\cos\phi} \] \[x-y=\frac{1-\sin\phi}{\cos\phi}-\frac{1+\cos\phi}{\sin\phi}=\frac{\sin\phi-\sin^2\phi-\cos\phi-\cos^2\phi}{\sin\phi\cos\phi} \] \[x-y=\frac{\sin\phi-\cos\phi-1}{\sin\phi\cos\phi}=-\frac{1+\cos\phi-\sin\phi}{\sin\phi\cos\phi}=-(xy+1) \] \[xy+x-y+1=0 \]

  5. Exercise 25

    If θ\displaystyle \theta lies in the first quadrant and cos⁡θ=817\displaystyle \cos \theta=\frac{8}{17}, then find the value of cos⁡(30∘+θ)+cos⁡(45∘−θ)+cos⁡(120∘−θ)\displaystyle \cos \left(30^{\circ}+\theta\right)+\cos \left(45^{\circ}-\theta\right)+\cos \left(120^{\circ}-\theta\right).

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    NCERT’s answer
    \(\displaystyle \frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)\)
    \[\sin\theta = \sqrt{1-\tfrac{64}{289}} = \tfrac{15}{17} \quad (\theta \text{ in the first quadrant}) \] \[\cos(30^\circ+\theta) = \tfrac{\sqrt3}{2}\cdot\tfrac{8}{17} - \tfrac12\cdot\tfrac{15}{17} = \tfrac{8\sqrt3-15}{34} \] \[\cos(45^\circ-\theta) = \tfrac{1}{\sqrt2}\left(\tfrac{8}{17}+\tfrac{15}{17}\right) = \tfrac{23\sqrt2}{34} \] \[\cos(120^\circ-\theta) = -\tfrac12\cdot\tfrac{8}{17} + \tfrac{\sqrt3}{2}\cdot\tfrac{15}{17} = \tfrac{15\sqrt3-8}{34} \] \[\text{Sum} = \frac{(8\sqrt3-15)+23\sqrt2+(15\sqrt3-8)}{34} = \frac{23\left(\sqrt3+\sqrt2-1\right)}{34} \] Answer: \(\displaystyle \dfrac{23\left(\sqrt3+\sqrt2-1\right)}{34}\)

  6. Exercise 26

    Find the value of the expression cos⁡4π8+cos⁡43π8+cos⁡45π8+cos⁡47π8\displaystyle \cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} [Hint: Simplify the expression to 2(cos⁡4π8+cos⁡43π8)\displaystyle 2\left(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}\right) =2[(cos⁡2π8+cos⁡23π8)2−2cos⁡2π8cos⁡23π8]\displaystyle =2\left[\left(\cos ^2 \frac{\pi}{8}+\cos ^2 \frac{3 \pi}{8}\right)^2-2 \cos ^2 \frac{\pi}{8} \cos ^2 \frac{3 \pi}{8}\right]

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    NCERT’s answer
    \(\displaystyle \frac{3}{2}\)
    \[\cos\tfrac{5\pi}{8} = -\cos\tfrac{3\pi}{8}, \qquad \cos\tfrac{7\pi}{8} = -\cos\tfrac{\pi}{8} \] \[E = 2\left(\cos^4\tfrac{\pi}{8}+\cos^4\tfrac{3\pi}{8}\right) = 2\left[\left(\cos^2\tfrac{\pi}{8}+\cos^2\tfrac{3\pi}{8}\right)^2 - 2\cos^2\tfrac{\pi}{8}\cos^2\tfrac{3\pi}{8}\right] \] \[\cos\tfrac{3\pi}{8} = \sin\tfrac{\pi}{8} \;\Rightarrow\; \cos^2\tfrac{\pi}{8}+\cos^2\tfrac{3\pi}{8} = 1 \] \[\cos^2\tfrac{\pi}{8}\cos^2\tfrac{3\pi}{8} = \left(\sin\tfrac{\pi}{8}\cos\tfrac{\pi}{8}\right)^2 = \left(\tfrac12\sin\tfrac{\pi}{4}\right)^2 = \tfrac18 \] \[E = 2\left[1 - 2\cdot\tfrac18\right] = \tfrac32 \] Answer: \(\displaystyle \dfrac32\)

  7. Exercise 27

    Find the general solution of the equation 5cos⁡2θ+7sin⁡2θ−6=0\displaystyle 5 \cos ^2 \theta+7 \sin ^2 \theta-6=0

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    NCERT’s answer
    \(\displaystyle n \pi \pm \frac{\pi}{4}\)
    \[5\left(\cos^2\theta+\sin^2\theta\right) + 2\sin^2\theta - 6 = 0 \] \[5 + 2\sin^2\theta - 6 = 0 \] \[\sin^2\theta = \tfrac12 = \sin^2\tfrac{\pi}{4} \] \[\theta = n\pi \pm \tfrac{\pi}{4}, \quad n\in\mathbb{Z} \] Answer: \(\displaystyle \theta = n\pi \pm \dfrac{\pi}{4}\), \(\displaystyle n\in\mathbb{Z}\)

  8. Exercise 28

    Find the general solution of the equation sin⁡x−3sin⁡2x+sin⁡3x=cos⁡x−3cos⁡2x+cos⁡3x\displaystyle \sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x

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    \[(\sin x+\sin 3x) - 3\sin 2x = (\cos x+\cos 3x) - 3\cos 2x \] \[2\sin 2x\cos x - 3\sin 2x = 2\cos 2x\cos x - 3\cos 2x \] \[\sin 2x\,(2\cos x-3) = \cos 2x\,(2\cos x-3) \] \[2\cos x - 3 \le -1 \ne 0 \;\Rightarrow\; \sin 2x = \cos 2x \] \[\tan 2x = 1 = \tan\tfrac{\pi}{4} \] \[2x = n\pi + \tfrac{\pi}{4} \] Answer: \(\displaystyle x = \dfrac{n\pi}{2} + \dfrac{\pi}{8}\), \(\displaystyle n\in\mathbb{Z}\)

  9. Exercise 29

    Find the general solution of the equation (3−1)cos⁡θ+(3+1)sin⁡θ=2\displaystyle (\sqrt{3}-1) \cos \theta+(\sqrt{3}+1) \sin \theta=2 [Hint: Put 3−1=rsin⁡α,3+1=rcos⁡α\displaystyle \sqrt{3}-1=r \sin \alpha, \sqrt{3}+1=r \cos \alpha which gives tan⁡α=tan⁡(π4−π6)\displaystyle \tan \alpha=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right) ⇒α=π12\displaystyle \Rightarrow \alpha=\frac{\pi}{12} ]

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[\sqrt3-1 = r\sin\alpha, \qquad \sqrt3+1 = r\cos\alpha \] \[r^2 = (\sqrt3-1)^2+(\sqrt3+1)^2 = 8 \;\Rightarrow\; r = 2\sqrt2 \] \[\tan\alpha = \frac{\sqrt3-1}{\sqrt3+1} = \frac{1-\tfrac{1}{\sqrt3}}{1+\tfrac{1}{\sqrt3}} = \tan\left(\tfrac{\pi}{4}-\tfrac{\pi}{6}\right) \;\Rightarrow\; \alpha = \tfrac{\pi}{12} \] \[r\sin\alpha\cos\theta + r\cos\alpha\sin\theta = 2 \] \[2\sqrt2\,\sin\left(\theta+\tfrac{\pi}{12}\right) = 2 \] \[\sin\left(\theta+\tfrac{\pi}{12}\right) = \tfrac{1}{\sqrt2} = \sin\tfrac{\pi}{4} \] \[\theta+\tfrac{\pi}{12} = n\pi + (-1)^n\tfrac{\pi}{4} \] Answer: \(\displaystyle \theta = n\pi + (-1)^n\dfrac{\pi}{4} - \dfrac{\pi}{12}\), \(\displaystyle n\in\mathbb{Z}\)
  10. Choose the correct answer from the given four options in the Exercises $\displaystyle 30$ to $\displaystyle 59$ (M.C.Q.).

    Exercise 30

    If sin⁡θ+cosec⁡θ=2\displaystyle \sin \theta+\operatorname{cosec} \theta=2, then sin⁡2θ+cosec⁡2θ\displaystyle \sin ^2 \theta+\operatorname{cosec}^2 \theta is equal to
    (A)
    1\displaystyle 1 (B) 4\displaystyle 4 (C) 2\displaystyle 2 (D) None of these

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    NCERT’s answer
    C
    (C) \(\displaystyle 2\) \[\sin^2\theta+\operatorname{cosec}^2\theta = (\sin\theta+\operatorname{cosec}\theta)^2 - 2\sin\theta\operatorname{cosec}\theta \] \[= 2^2 - 2 = 2 \]