SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Trigonometric Functions

52 questions · 52 still being checked

EXERCISE 3.3 11–20 (part 4 of 6)

  1. Prove the following:

    Exercise 11

    cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡x\displaystyle \cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x

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    Difference of cosines into a product.Rule: \(\displaystyle \cos A-\cos B=-2\sin\dfrac{A+B}{2}\,\sin\dfrac{A-B}{2}\).Take \(\displaystyle A=\frac{3\pi}{4}+x\) and \(\displaystyle B=\frac{3\pi}{4}-x\), so \(\displaystyle \dfrac{A+B}{2}=\dfrac{3\pi}{4}\) and \(\displaystyle \dfrac{A-B}{2}=x\):\[\text{LHS}=-2\sin\frac{3\pi}{4}\,\sin x\]Since \(\displaystyle \frac{3\pi}{4}=\pi-\frac{\pi}{4}\) is in the second quadrant, \(\displaystyle \sin\frac{3\pi}{4}=\frac{1}{\sqrt{2}}\), so\[\text{LHS}=-2\cdot\frac{1}{\sqrt{2}}\sin x=-\sqrt{2}\,\sin x=\text{RHS}\]Hence \(\displaystyle \cos\!\left(\dfrac{3\pi}{4}+x\right)-\cos\!\left(\dfrac{3\pi}{4}-x\right)=-\sqrt{2}\sin x\).

  2. Exercise 12

    sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x\displaystyle \sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x

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    Difference of squares of sines.First establish the rule \(\displaystyle \sin^{2}A-\sin^{2}B=\sin(A+B)\sin(A-B)\). Using \(\displaystyle \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}\),\[\sin^{2}A-\sin^{2}B=\frac{1-\cos 2A}{2}-\frac{1-\cos 2B}{2}=\frac{\cos 2B-\cos 2A}{2}\]and \(\displaystyle \cos 2B-\cos 2A=2\sin(A+B)\sin(A-B)\), so \(\displaystyle \sin^{2}A-\sin^{2}B=\sin(A+B)\sin(A-B)\).Apply it with \(\displaystyle A=6x,\;B=4x\):\[\text{LHS}=\sin(6x+4x)\sin(6x-4x)=\sin 10x\,\sin 2x=\sin 2x\,\sin 10x=\text{RHS}\]Hence \(\displaystyle \sin^{2}6x-\sin^{2}4x=\sin 2x\,\sin 10x\).

  3. Exercise 13

    cos⁡22x−cos⁡26x=sin⁡4xsin⁡8x\displaystyle \cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x

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    Difference of squares of cosines.Using \(\displaystyle \cos^{2}\theta=\dfrac{1+\cos 2\theta}{2}\),\[\cos^{2}2x-\cos^{2}6x=\frac{1+\cos 4x}{2}-\frac{1+\cos 12x}{2}=\frac{\cos 4x-\cos 12x}{2}\]Now apply \(\displaystyle \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) with \(\displaystyle A=4x,\;B=12x\):\[\cos 4x-\cos 12x=-2\sin 8x\,\sin(-4x)=2\sin 8x\,\sin 4x\]Therefore\[\text{LHS}=\frac{2\sin 8x\,\sin 4x}{2}=\sin 4x\,\sin 8x=\text{RHS}\]Hence \(\displaystyle \cos^{2}2x-\cos^{2}6x=\sin 4x\,\sin 8x\).

  4. Exercise 14

    sin⁡2x+2sin⁡4x+sin⁡6x=4cos⁡2xsin⁡4x\displaystyle \sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x

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    Sum of sines into a product.Group the two outer terms and use \(\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) with \(\displaystyle A=6x,\;B=2x\):\[\sin 2x+\sin 6x=2\sin 4x\,\cos 2x\]So\[\text{LHS}=2\sin 4x\,\cos 2x+2\sin 4x=2\sin 4x\,(1+\cos 2x)\]Since \(\displaystyle 1+\cos 2x=2\cos^{2}x\),\[\text{LHS}=2\sin 4x\cdot 2\cos^{2}x=4\cos^{2}x\,\sin 4x=\text{RHS}\]Hence \(\displaystyle \sin 2x+2\sin 4x+\sin 6x=4\cos^{2}x\,\sin 4x\).

  5. Exercise 15

    cot⁡4x(sin⁡5x+sin⁡3x)=cot⁡x(sin⁡5x−sin⁡3x)\displaystyle \cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)

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    Sum and difference of sines into products.Rules: \(\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) and \(\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\), here with \(\displaystyle A=5x,\;B=3x\).\[\sin 5x+\sin 3x=2\sin 4x\,\cos x,\qquad \sin 5x-\sin 3x=2\cos 4x\,\sin x\]Left side:\[\text{LHS}=\frac{\cos 4x}{\sin 4x}\cdot 2\sin 4x\,\cos x=2\cos 4x\,\cos x\]Right side:\[\text{RHS}=\frac{\cos x}{\sin x}\cdot 2\cos 4x\,\sin x=2\cos 4x\,\cos x\]Both sides reduce to the same expression.Hence \(\displaystyle \cot 4x\,(\sin 5x+\sin 3x)=\cot x\,(\sin 5x-\sin 3x)\).

  6. Exercise 16

    cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x\displaystyle \frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}

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    Sum-to-product on numerator and denominator.Numerator, by \(\displaystyle \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) with \(\displaystyle A=9x,\;B=5x\):\[\cos 9x-\cos 5x=-2\sin 7x\,\sin 2x\]Denominator, by \(\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) with \(\displaystyle A=17x,\;B=3x\):\[\sin 17x-\sin 3x=2\cos 10x\,\sin 7x\]Dividing and cancelling \(\displaystyle 2\sin 7x\),\[\text{LHS}=\frac{-2\sin 7x\,\sin 2x}{2\cos 10x\,\sin 7x}=-\frac{\sin 2x}{\cos 10x}=\text{RHS}\]Hence \(\displaystyle \dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}\).

  7. Exercise 17

    sin⁡5x+sin⁡3xcos⁡5x+cos⁡3x=tan⁡4x\displaystyle \frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x

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    Sum-to-product on numerator and denominator.With \(\displaystyle A=5x,\;B=3x\), so \(\displaystyle \dfrac{A+B}{2}=4x\) and \(\displaystyle \dfrac{A-B}{2}=x\):\[\sin 5x+\sin 3x=2\sin 4x\,\cos x,\qquad \cos 5x+\cos 3x=2\cos 4x\,\cos x\]Dividing and cancelling \(\displaystyle 2\cos x\),\[\text{LHS}=\frac{2\sin 4x\,\cos x}{2\cos 4x\,\cos x}=\frac{\sin 4x}{\cos 4x}=\tan 4x=\text{RHS}\]Hence \(\displaystyle \dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x\).

  8. Exercise 18

    sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\displaystyle \frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}

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    Sum-to-product on numerator and denominator.Rules: \(\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) and \(\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\), here with \(\displaystyle A=x,\;B=y\).\[\sin x-\sin y=2\cos\frac{x+y}{2}\,\sin\frac{x-y}{2}\] \[\cos x+\cos y=2\cos\frac{x+y}{2}\,\cos\frac{x-y}{2}\]Dividing and cancelling \(\displaystyle 2\cos\dfrac{x+y}{2}\),\[\text{LHS}=\frac{\sin\dfrac{x-y}{2}}{\cos\dfrac{x-y}{2}}=\tan\frac{x-y}{2}=\text{RHS}\]Hence \(\displaystyle \dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan\dfrac{x-y}{2}\).

  9. Exercise 19

    sin⁡x+sin⁡3xcos⁡x+cos⁡3x=tan⁡2x\displaystyle \frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x

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    Sum-to-product on numerator and denominator.With \(\displaystyle A=3x,\;B=x\), so \(\displaystyle \dfrac{A+B}{2}=2x\) and \(\displaystyle \dfrac{A-B}{2}=x\):\[\sin x+\sin 3x=2\sin 2x\,\cos x,\qquad \cos x+\cos 3x=2\cos 2x\,\cos x\]Dividing and cancelling \(\displaystyle 2\cos x\),\[\text{LHS}=\frac{2\sin 2x\,\cos x}{2\cos 2x\,\cos x}=\frac{\sin 2x}{\cos 2x}=\tan 2x=\text{RHS}\]Hence \(\displaystyle \dfrac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x\).

  10. Exercise 20

    sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\displaystyle \frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x

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    Sum-to-product on the numerator, double angle on the denominator.Use \(\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) with \(\displaystyle A=x,\;B=3x\):\[\sin x-\sin 3x=2\cos 2x\,\sin(-x)=-2\sin x\cos 2x . \]For the denominator use \(\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x\), so\[\sin^{2}x-\cos^{2}x=-(\cos^{2}x-\sin^{2}x)=-\cos 2x . \]Therefore\[\frac{\sin x-\sin 3x}{\sin^{2}x-\cos^{2}x} =\frac{-2\sin x\cos 2x}{-\cos 2x} =2\sin x , \]the cancellation being valid wherever \(\displaystyle \cos 2x\neq 0\), which is exactly where the left side is defined.Hence \(\displaystyle \dfrac{\sin x-\sin 3x}{\sin^{2}x-\cos^{2}x}=2\sin x\).