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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

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EXERCISE 3.3 61–70 (part 7 of 8)

  1. Fill in the blanks in Exercises $\displaystyle 60$ to $\displaystyle 67$ :

    Exercise 61

    If k=sin⁡(π18)sin⁡(5π18)sin⁡(7π18)\displaystyle k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right), then the numerical value of k\displaystyle k is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \frac{1}{8}\)
    \(\displaystyle \dfrac{1}{8}\)\[\frac{\pi}{18}=10^\circ,\quad \frac{5\pi}{18}=50^\circ,\quad \frac{7\pi}{18}=70^\circ \] \[k = \sin 10^\circ\sin 50^\circ\sin 70^\circ = \cos 80^\circ\cos 40^\circ\cos 20^\circ \] \[8\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ = 4\sin 40^\circ\cos 40^\circ\cos 80^\circ = 2\sin 80^\circ\cos 80^\circ = \sin 160^\circ = \sin 20^\circ \] \[k = \frac{\sin 20^\circ}{8\sin 20^\circ} = \frac{1}{8} \]Answer: \(\displaystyle \dfrac{1}{8}\)

  2. Exercise 62

    If tan⁡A=1−cos⁡Bsin⁡B\displaystyle \tan \mathrm{A}=\frac{1-\cos \mathrm{B}}{\sin \mathrm{B}}, then tan⁡2 A=\displaystyle \tan 2 \mathrm{~A}= ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \tan \beta\)
    \(\displaystyle \tan B\)\[\tan A = \frac{1-\cos B}{\sin B} = \frac{2\sin^2\dfrac{B}{2}}{2\sin\dfrac{B}{2}\cos\dfrac{B}{2}} = \tan\frac{B}{2} \] \[\tan 2A = \frac{2\tan A}{1-\tan^2 A} = \frac{2\tan\dfrac{B}{2}}{1-\tan^2\dfrac{B}{2}} = \tan B \]Answer: \(\displaystyle \tan B\)

  3. Exercise 63

    If sin⁡x+cos⁡x=a\displaystyle \sin x+\cos x=a, then
    (i)
    sin⁡6x+cos⁡6x=\displaystyle \sin ^6 x+\cos ^6 x= ____\displaystyle \_\_\_\_
    (ii)
    ∣sin⁡x−cos⁡x∣=\displaystyle |\sin x-\cos x|= ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\left[4-3\left(a^2-1\right)^2\right], \sqrt{2-a^2}\)
    (i) \(\displaystyle \dfrac{1+6a^2-3a^4}{4}\), (ii) \(\displaystyle \sqrt{2-a^2}\)
    \[(\sin x+\cos x)^2 = a^2 \;\Rightarrow\; 1+2\sin x\cos x = a^2 \;\Rightarrow\; \sin x\cos x = \frac{a^2-1}{2} \]
    (i)
    \[\sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x+\cos^2x) = 1-3\sin^2x\cos^2x \]
    \[= 1-\frac{3(a^2-1)^2}{4} = \frac{1+6a^2-3a^4}{4} \]
    (ii)
    \[(\sin x-\cos x)^2 = 1-2\sin x\cos x = 1-(a^2-1) = 2-a^2 \]
    \[|\sin x-\cos x| = \sqrt{2-a^2} \]
    Answer: (i) \(\displaystyle \dfrac{1+6a^2-3a^4}{4}\); (ii) \(\displaystyle \sqrt{2-a^2}\)

  4. Exercise 64

    In a triangle ABC with ∠C=90∘\displaystyle \angle \mathrm{C}=90^{\circ} the equation whose roots are tan⁡A\displaystyle \tan \mathrm{A} and tan⁡B\displaystyle \tan \mathrm{B} is ____\displaystyle \_\_\_\_. [Hint: A+B=90∘⇒tan⁡Atan⁡B=1\displaystyle \mathrm{A}+\mathrm{B}=90^{\circ} \Rightarrow \tan \mathrm{A} \tan \mathrm{B}=1 and tan⁡A+tan⁡B=2sin⁡2 A\displaystyle \tan \mathrm{A}+\tan \mathrm{B}=\frac{2}{\sin 2 \mathrm{~A}} ]

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    NCERT’s answer
    \(\displaystyle x^2-\frac{2}{\sin 2 \mathrm{~A}} x+1\)
    \(\displaystyle x^2-\dfrac{2}{\sin 2A}\,x+1=0\)\[A+B=90^\circ \;\Rightarrow\; \tan B=\cot A \;\Rightarrow\; \tan A\tan B = 1 \] \[\tan A+\tan B = \tan A+\cot A = \frac{\sin^2A+\cos^2A}{\sin A\cos A} = \frac{2}{\sin 2A} \] \[x^2-(\tan A+\tan B)\,x+\tan A\tan B = 0 \]Answer: \(\displaystyle x^2-\dfrac{2}{\sin 2A}\,x+1=0\)

  5. Exercise 65

    3(sin⁡x−cos⁡x)4+6(sin⁡x+cos⁡x)2+4(sin⁡6x+cos⁡6x)=\displaystyle 3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4\left(\sin ^6 x+\cos ^6 x\right)= ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 13$
    $\displaystyle 13$Put \(\displaystyle p=\sin x\cos x\). \[(\sin x-\cos x)^2=1-2p,\qquad (\sin x+\cos x)^2=1+2p \] \[\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3p^2 \] \[3(1-2p)^2+6(1+2p)+4(1-3p^2)=3-12p+12p^2+6+12p+4-12p^2 \] \[=13 \]

  6. Exercise 66

    Given x>0\displaystyle x>0, the values of f(x)=−3cos⁡3+x+x2\displaystyle f(x)=-3 \cos \sqrt{3+x+x^2} lie in the interval ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    [-$\displaystyle 3$, $\displaystyle 3$]
    \(\displaystyle [-3,\,3]\)For \(\displaystyle x>0\) the argument is continuous and unbounded, and \[\sqrt{3+x+x^2}>\sqrt3\approx1.73 \] \[\sqrt{3+x+x^2}=\pi \;\Rightarrow\; \cos\sqrt{3+x+x^2}=-1 \;\Rightarrow\; f(x)=3 \] \[\sqrt{3+x+x^2}=2\pi \;\Rightarrow\; \cos\sqrt{3+x+x^2}=1 \;\Rightarrow\; f(x)=-3 \] Both values occur (\(\displaystyle \pi>\sqrt3\)), and always \[-1\le\cos\sqrt{3+x+x^2}\le1 \;\Rightarrow\; -3\le f(x)\le3 \] NCERT_Solution_Class11_Maths_Exemplar_Ch3_Ex3-3_Q66

  7. Exercise 67

    The maximum distance of a point on the graph of the function y=3sin⁡x+cos⁡x\displaystyle y=\sqrt{3} \sin x+\cos x from x\displaystyle x-axis is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 2$
    $\displaystyle 2$\[y=\sqrt3\sin x+\cos x=2\left(\tfrac{\sqrt3}{2}\sin x+\tfrac12\cos x\right) \] \[y=2\sin\left(x+\tfrac{\pi}{6}\right) \] \[-2\le y\le2,\qquad y=2 \text{ at } x=\tfrac{\pi}{3} \] Distance from the \(\displaystyle x\)-axis is \(\displaystyle |y|\): \[|y|_{\max}=2 \] NCERT_Solution_Class11_Maths_Exemplar_Ch3_Ex3-3_Q67
  8. In each of the Exercises $\displaystyle 68$ to $\displaystyle 75$, state whether the statements is True or False? Also give justification.

    Exercise 68

    If tan⁡A=1−cos⁡Bsin⁡B\displaystyle \tan \mathrm{A}=\frac{1-\cos \mathrm{B}}{\sin \mathrm{B}}, then tan⁡2 A=tan⁡B\displaystyle \tan 2 \mathrm{~A}=\tan \mathrm{B}

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    NCERT’s answer
    True
    True\[\tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2\frac B2}{2\sin\frac B2\cos\frac B2}=\tan\frac B2 \] \[\tan2A=\frac{2\tan A}{1-\tan^2A}=\frac{2\tan\frac B2}{1-\tan^2\frac B2}=\tan B \]

  9. Exercise 69

    The equality sin⁡A+sin⁡2 A+sin⁡3 A=3\displaystyle \sin \mathrm{A}+\sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}=3 holds for some real value of A.

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    NCERT’s answer
    False
    FalseEach term is at most $\displaystyle 1$, so the sum is $\displaystyle 3$ only if all three equal 1. But \[\sin A=1 \;\Rightarrow\; A=\tfrac\pi2+2n\pi \] \[\sin2A=\sin(\pi+4n\pi)=0\neq1 \] No real \(\displaystyle A\) works.

  10. Exercise 70

    sin⁡10∘\displaystyle \sin 10^{\circ} is greater than cos⁡10∘\displaystyle \cos 10^{\circ}.

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    NCERT’s answer
    False
    False\[\cos10^\circ=\sin80^\circ \] \[\sin10^\circ<\sin80^\circ \quad(\sin \text{ increases on } [0^\circ,90^\circ]) \] \[\sin10^\circ\approx0.17<0.98\approx\cos10^\circ \]