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Accedmychevron_right11thchevron_rightmathchevron_rightTrigonometric Identitieschevron_rightExercise 10.2

Questions

  1. Question 1

    Without using table find the values of the following:

    (i)

    sin⁡15∘\sin 15^\circ

    (ii)

    cos⁡15∘\cos 15^\circ

    (iii)

    tan⁡15∘\tan 15^\circ

    (iv)

    sin⁡105∘\sin 105^\circ

    (v)

    cos⁡105∘\cos 105^\circ

    (vi)

    tan⁡105∘\tan 105^\circ

    SolutionTheory
  2. Question 2

    Prove that:

    (i)

    sin⁡(45∘+α)=12(sin⁡α+cos⁡α)\sin(45^\circ + \alpha) = \dfrac{1}{\sqrt{2}}(\sin\alpha + \cos\alpha)

    (ii)

    cos⁡(α+45∘)=12(cos⁡α−sin⁡α)\cos(\alpha + 45^\circ) = \dfrac{1}{\sqrt{2}}(\cos\alpha - \sin\alpha)

    SolutionTheory
  3. Question 3

    Prove that:

    (i)

    tan⁡(45∘+A)tan⁡(45∘−A)=1\tan(45^\circ + A)\tan(45^\circ - A) = 1

    (ii)

    tan⁡(π4−θ)+tan⁡(3π4+θ)=0\tan\left(\dfrac{\pi}{4} - \theta\right) + \tan\left(\dfrac{3\pi}{4} + \theta\right) = 0

    (iii)

    sin⁡(θ+π6)+cos⁡(θ+π3)=cos⁡θ\sin\left(\theta + \dfrac{\pi}{6}\right) + \cos\left(\theta + \dfrac{\pi}{3}\right) = \cos\theta

    (iv)

    sin⁡θ−cos⁡θtan⁡θ2cos⁡θ+sin⁡θtan⁡θ2=tan⁡θ2\dfrac{\sin\theta - \cos\theta \tan\dfrac{\theta}{2}}{\cos\theta + \sin\theta \tan\dfrac{\theta}{2}} = \tan\dfrac{\theta}{2}

    (v)

    1−tan⁡θtan⁡ϕ1+tan⁡θtan⁡ϕ=cos⁡(θ+ϕ)cos⁡(θ−ϕ)\dfrac{1 - \tan\theta \tan\phi}{1 + \tan\theta \tan\phi} = \dfrac{\cos(\theta + \phi)}{\cos(\theta - \phi)}

    SolutionTheory
  4. Question 4

    Show that: cos⁡(α+β)cos⁡(α−β)=cos⁡2α−sin⁡2β=cos⁡2β−sin⁡2α\cos(\alpha + \beta)\cos(\alpha - \beta) = \cos^2\alpha - \sin^2\beta = \cos^2\beta - \sin^2\alpha.

    SolutionTheory
  5. Question 5

    Show that: sin⁡(α+β)+sin⁡(α−β)cos⁡(α+β)+cos⁡(α−β)=tan⁡α\dfrac{\sin(\alpha + \beta) + \sin(\alpha - \beta)}{\cos(\alpha + \beta) + \cos(\alpha - \beta)} = \tan\alpha.

    SolutionTheory
  6. Question 6

    Show that:

    (i)

    sin⁡2(α+β2)−sin⁡2(α−β2)=sin⁡2αsin⁡β\sin^2\left(\alpha + \dfrac{\beta}{2}\right) - \sin^2\left(\alpha - \dfrac{\beta}{2}\right) = \sin 2\alpha \sin\beta

    (ii)

    sin⁡2α+sin⁡2β+cos⁡2(α+β)+2sin⁡αsin⁡βcos⁡(α+β)=1\sin^2\alpha + \sin^2\beta + \cos^2(\alpha + \beta) + 2\sin\alpha\sin\beta\cos(\alpha + \beta) = 1

    SolutionTheory
  7. Question 7

    Show that:

    (i)

    cos⁡(α−β)=1+tan⁡αtan⁡βsec⁡αsec⁡β\cos(\alpha - \beta) = \dfrac{1 + \tan\alpha \tan\beta}{\sec\alpha \sec\beta}

    (ii)

    sin⁡(α+β)=1+cot⁡αtan⁡βcosec⁡αsec⁡β\sin(\alpha + \beta) = \dfrac{1 + \cot\alpha \tan\beta}{\operatorname{cosec}\alpha \sec\beta}

    (iii)

    cot⁡(α−β)=cot⁡αcot⁡β+1cot⁡β−cot⁡α\cot(\alpha - \beta) = \dfrac{\cot\alpha \cot\beta + 1}{\cot\beta - \cot\alpha}

    (iv)

    tan⁡α+tan⁡βtan⁡α−tan⁡β=sin⁡(α+β)sin⁡(α−β)\dfrac{\tan\alpha + \tan\beta}{\tan\alpha - \tan\beta} = \dfrac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)}

    (v)

    cot⁡(α+β)=cot⁡αcot⁡β−1cot⁡α+cot⁡β\cot(\alpha + \beta) = \dfrac{\cot\alpha \cot\beta - 1}{\cot\alpha + \cot\beta}

    SolutionTheory
  8. Question 8

    If sin⁡α=2425\sin\alpha = \dfrac{24}{25} and cos⁡β=2029\cos\beta = \dfrac{20}{29}, where 0<α<π20 < \alpha < \dfrac{\pi}{2} and 0<β<π20 < \beta < \dfrac{\pi}{2}. Show that sin⁡(α−β)=333725\sin(\alpha - \beta) = \dfrac{333}{725}.

    SolutionTheory
  9. Question 9

    If sin⁡α=−817\sin\alpha = -\dfrac{8}{17} and cos⁡β=−45\cos\beta = -\dfrac{4}{5} where 3π2<α<2π\dfrac{3\pi}{2} < \alpha < 2\pi and π<β<3π2\pi < \beta < \dfrac{3\pi}{2}. Find

    (i)

    sin⁡(α+β)\sin(\alpha + \beta)

    (ii)

    cos⁡(α+β)\cos(\alpha + \beta)

    (iii)

    tan⁡(α+β)\tan(\alpha + \beta)

    (iv)

    sin⁡(α−β)\sin(\alpha - \beta)

    (v)

    cos⁡(α−β)\cos(\alpha - \beta)

    (vi)

    tan⁡(α−β)\tan(\alpha - \beta)

    SolutionTheory
  10. Question 10

    Find sin⁡(α+β)\sin(\alpha + \beta) and cos⁡(α+β)\cos(\alpha + \beta), given that

    (i)

    tan⁡α=34\tan\alpha = \dfrac{3}{4}, cos⁡β=513\cos\beta = \dfrac{5}{13} and neither the terminal side of the angle of measure α\alpha nor that of β\beta is in the quadrant I.

    (ii)

    tan⁡α=−158\tan\alpha = -\dfrac{15}{8} and sin⁡β=−725\sin\beta = -\dfrac{7}{25} and neither the terminal side of the angle of measure α\alpha nor that of β\beta is in the quadrant IV.

    SolutionTheory
  11. Question 11

    Prove that: cos⁡19∘+sin⁡19∘cos⁡19∘−sin⁡19∘=tan⁡64∘\dfrac{\cos 19^\circ + \sin 19^\circ}{\cos 19^\circ - \sin 19^\circ} = \tan 64^\circ.

    SolutionTheory
  12. Question 12

    Prove that: cos⁡(60∘+θ)cos⁡(60∘−θ)+sin⁡(60∘+θ)sin⁡(60∘−θ)=cos⁡2θ\cos(60^\circ + \theta)\cos(60^\circ - \theta) + \sin(60^\circ + \theta)\sin(60^\circ - \theta) = \cos 2\theta.

    SolutionTheory
  13. Question 13

    If α,β,γ\alpha, \beta, \gamma are the angles of a triangle ABCABC, show that cot⁡α2+cot⁡β2+cot⁡γ2=cot⁡α2cot⁡β2cot⁡γ2\cot\dfrac{\alpha}{2} + \cot\dfrac{\beta}{2} + \cot\dfrac{\gamma}{2} = \cot\dfrac{\alpha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}.

    SolutionTheory
  14. Question 14

    If α+β+γ=180∘\alpha + \beta + \gamma = 180^\circ, show that: cot⁡αcot⁡β+cot⁡βcot⁡γ+cot⁡γcot⁡α=1\cot\alpha \cot\beta + \cot\beta \cot\gamma + \cot\gamma \cot\alpha = 1.

    SolutionTheory
  15. Question 15

    Express the following in the form rsin⁡(θ+ϕ)r\sin(\theta + \phi) or rsin⁡(θ−ϕ)r\sin(\theta - \phi) where terminal sides of the angles of measures θ\theta and ϕ\phi are in the first quadrant:

    (i)

    24sin⁡θ+7cos⁡θ24\sin\theta + 7\cos\theta

    (ii)

    12sin⁡θ−5cos⁡θ12\sin\theta - 5\cos\theta

    (iii)

    sin⁡θ−cos⁡θ\sin\theta - \cos\theta

    (iv)

    8sin⁡θ−6cos⁡θ8\sin\theta - 6\cos\theta

    (v)

    12sin⁡θ+32cos⁡θ\dfrac{1}{2}\sin\theta + \dfrac{\sqrt{3}}{2}\cos\theta

    (vi)

    13sin⁡θ−84cos⁡θ13\sin\theta - 84\cos\theta

    SolutionTheory
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