Pembahasan Soal Trigonometri UN SMA (1)


  1. Nilai dari cos 40° + cos 80° + cos 160° = …

    A. -1/2 \sqrt{2}

    B. -1/2

    C. 0

    D. 1/2

    E. 1/2 \sqrt{2}

    PEMBAHASAN :

    NOTE :

    Cos (\alpha + \beta) = cos \alpha cos \betasin \alpha sin \beta

    Cos (\alpha-\beta) = cos \alpha cos \beta + sin \alpha sin \beta

    Jumlahkan kedua persamaan tersebut sehingga diperoleh :

    Cos (\alpha + \beta) + Cos (\alpha-\beta) = 2cos \alpha cos \beta

    Cos \frac12(\alpha + \beta) + Cos \frac12(\alpha-\beta) = cos \alpha cos \beta

    cos 40° + cos 80° + cos 160° = (cos 40° + cos 80°) + cos 160°

    = (cos ½(a + b) + cos ½(a – b)) + cos 1600

        ½(a + b) = 400    ½(a – b) = 800

    a + b = 1600      a – b = 800

    a + b = 1600

    a – b = 800 +

    2a = 2400

    a = 1200 \Rightarrow b = 400

    = (cos a cos b) + cos 1600

    = cos 1200 cos 400 + cos 1600

    = -1/2 cos 400 + cos 1600

    = -1/2 cos 400 + cos (1200 + 400)

    =

    =

    JAWABAN :

  2. Nilai sin 105° + cos 15° = …

    A. 1/2 (-\sqrt{2}-\sqrt{2})

    B. 1/2 (\sqrt{3}-\sqrt{2})

    C. 1/2 (\sqrt{6}-\sqrt{2})

    D. 1/2 (\sqrt{3}+\sqrt{2})

    E. 1/2 (\sqrt{6}+\sqrt{2})

    PEMBAHASAN :

    NOTE :

    Sin (\alpha+\beta) = sin \alpha cos \beta + cos \alpha sin \beta

    Sin (600 + 450) = sin 600 cos 450 + cos 600 sin 450

    = \frac{1}{2}\sqrt{3} \frac{1}{2}\sqrt{2} + \frac{1}{2} \frac{1}{2}\sqrt{2}

    = \frac{1}{4}\sqrt{6} + \frac{1}{4}\sqrt{2}

    Cos (\alpha - \beta) = cos \alpha cos \beta + sin \alpha sin \beta

    Cos (600 – 450) = cos 600 cos 450 + sin 600 sin 450

    = \frac{1}{2} \frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{3} \frac{1}{2}\sqrt{2}

    = \frac{1}{4}\sqrt{2} + \frac{1}{4}\sqrt{6}

    sin 105° + cos 15° = \frac{1}{4}\sqrt{6} + \frac{1}{4}\sqrt{2} + \frac{1}{4}\sqrt{2} + \frac{1}{4}\sqrt{6}

    = \frac{1}{2}\sqrt{6} + \frac{1}{2}\sqrt{2}

    = \frac{1}{2}(\sqrt{6}+\sqrt{2})

    JAWABAN : E

  3. Nilai dari tan 165° = …

    A. 1 – \sqrt{3}

    B. -1 + \sqrt{3}

    C. -2 – \sqrt{3}

    D. 2 – \sqrt{3}

    E. 2 + \sqrt{3}

    PEMBAHASAN :

    tan 165° = tan (1800 – 15°)

    = \frac{tan180^0+tan15^0}{1-tan180^0.tan15^0}

    = \frac{0+tan15^0}{1-0.tan15^0}

    = tan 150

    tan 150 = tan (600 – 450)

    = \frac{tan60^0+tan45^0}{1-tan60^0.tan45^0}

    = \frac{\frac{1}{3}\sqrt{3}+1}{1-\frac{1}{3}\sqrt{3}.1}

    = \frac{\frac{1}{3}\sqrt{3}+1}{1-\frac{1}{3}\sqrt{3}}

    = \frac{1+\frac{1}{3}\sqrt{3}}{1-\frac{1}{3}\sqrt{3}} x \frac{1+\frac{1}{3}\sqrt{3}}{1+\frac{1}{3}\sqrt{3}}

    = \frac{1+\frac{2}{3}\sqrt{3}+\frac{1}{3}}{1-\frac{1}{3}}

    = \frac{\frac{4}{3}+\frac{2}{3}\sqrt{3}}{\frac{2}{3}}

    = 2 + \sqrt{3}

    JAWABAN : E

  4. Diketahui persamaan cos 2x + cos x = 0, untuk 0 < x < π nilai x yang memenuhi adalah …

    A. \frac{\pi}{6} dan \frac{\pi}{2}

    B. \frac{\pi}{2} dan \pi

    C. \frac{\pi}{3} dan \frac{\pi}{2}

    D. \frac{\pi}{3} dan \pi

    E. \frac{\pi}{6} dan \frac{\pi}{3}

    PEMBAHASAN :

    cos 2x + cos x = 0

    2cos2x – 1 + cos x = 0 (INGAT : cos 2x = 2cos2x – 1, baca DISINI)

    (2 cos x – 1)(cos x + 1)

    cos x = 1/2 atau cos x = -1

    cos x = \frac{\pi}{3} atau cos x = \pi

    JAWABAN : D

  5. Diketahui cos (x – y) = 4/5 dan sin x.sin y = 3/10. Nilai tan x.tan y = …

    A. -5/3

    B. -4/3

    C. -3/5

    D. 3/5

    E. 5/3

    PEMBAHASAN :

    cos (x – y) = cos x cos y + sin x sin y

    4/5 = cos x cos y + 3/10

    4/5 – 3/10 = cos x cos y

    1/2 = cos x cos y

    tan x.tan y = (sin x sin y)/(cos x cos y)

    = (3/10) / (1/2)

    = 3/5

    JAWABAN : D

  6. Nilai sin 15° = …

    A. \frac{1}{2} \sqrt{2-\sqrt{2}}

    B. \frac{1}{2} (\sqrt{2}-\sqrt{6}

    C. \frac{1}{4} (\sqrt{2}+1)

    D. \frac{1}{4} (\sqrt{6}-\sqrt{2})

    E. \frac{1}{2} (\sqrt{2}+\sqrt{6}

    PEMBAHASAN :

    Sin (600 – 450) = sin 600 cos 450 – cos 600 sin 450

    = \frac{1}{2}\sqrt{3} \frac{1}{2}\sqrt{2}\frac{1}{2} \frac{1}{2}\sqrt{2}

    = \frac{1}{4}\sqrt{6}\frac{1}{4}\sqrt{2}

    = \frac{1}{4}(\sqrt{6}-\sqrt{2})

    JAWABAN : D

  7. Diketahui sin x = 8/10, 0 < x < 90°. Nilai cos 3x = …

    A. -18/25

    B. -84/125

    C. -42/125

    D. 6/25

    E. -12/25

    PEMBAHASAN :

    sin x = 8/10 \Rightarrow cos x = 6/10

    cos 3x = cos (2x + x)

    = (cos 2x)(cos x) – (sin 2x)(sin x)

    = cos (x + x)(cos x) – (sin (x + x))(sin x)

    = (cos2 x – sin2 x)(cos x) – (sin x cos x + cos x sin x)(sin x)

    = ((3/5)2 – (4/5)2)(3/5) – (4/5.3/5 + 3/5.4/5)(4/5)

    = (9/25 – 16/25)(3/5) – (12/25 + 12/25)(4/5)

    = (-7/25)(3/5) – (24/25)(4/5)

    = (-21/125) – (96/125)

    = – 117/125

    JAWABAN :

  8. Bentuk \frac{2tan x}{1+tan^2x} ekivalen dengan …

    A. 2 sin x

    B. sin 2x

    C. 2 cos 2x

    D. cos 2x

    E. tan 2x

    PEMBAHASAN :

    INGAT :

    tan (\alpha+\beta) = \frac{tan \alpha + tan \beta}{1+tan \alpha.tan \beta} (baca DISINI)

    misal x = \alpha = \beta, maka :

    = \frac{tan (x) + tan (x)}{1+tan (x).tan (x)}

    = \frac{2.tan (x)}{1+tan^2x}

    = tan (x + x)

    = tan 2x

    JAWABAN : E

NOTE : silahkan dikoreksi dan berikan komentar jika ada kesalahan atau masih ada keambiguan dalam penyelesaian soal-soal ini.

8 comments on “Pembahasan Soal Trigonometri UN SMA (1)

  1. lohlahloh ?? cos 160 kenapa= cos(140+20) ??? harusnya kan pakai sudut 90 / 180 / 270 / 360 bray …jadi cos160=cos(180-20)=cos 20 …..gimana sih sampeyan 😀

  2. mas numpang nanya nih…Tentukan nilai x yang memnuhi persamaan sin 3x + cos 3x = 0 untuk 180 drjt < x < 360 drjt….makasih ya..

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