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cos(alpha+beta) dari tangen sudut

Trigonometri Matematika · SMA medium

Diketahui tan⁡α=34\tan\alpha = \dfrac{3}{4} dan tan⁡β=512\tan\beta = \dfrac{5}{12}; α\alpha dan β\beta sudut lancip. Maka nilai cos⁡(α+β)=…\cos(\alpha+\beta) = \ldots

  1. A
    6465\dfrac{64}{65}
  2. B
    6365\dfrac{63}{65}
  3. C
    3665\dfrac{36}{65}
  4. D
    3365\dfrac{33}{65}
    ✓
  5. E
    3065\dfrac{30}{65}

Kunci jawaban

D

Penjelasan singkat

α:sin⁡35,cos⁡45\alpha:\sin\tfrac35,\cos\tfrac45; β:sin⁡513,cos⁡1213\beta:\sin\tfrac5{13},\cos\tfrac{12}{13}. cos⁡(α+β)=45⋅1213−35⋅513=3365\cos(\alpha+\beta)=\tfrac45\cdot\tfrac{12}{13}-\tfrac35\cdot\tfrac5{13}=\tfrac{33}{65}.

(Jawaban dihitung & diverifikasi dengan Python/SymPy.)

Dari tan⁡α=34\tan\alpha=\tfrac34: sin⁡α=35,cos⁡α=45\sin\alpha=\tfrac35,\cos\alpha=\tfrac45. Dari tan⁡β=512\tan\beta=\tfrac5{12}: sin⁡β=513,cos⁡β=1213\sin\beta=\tfrac5{13},\cos\beta=\tfrac{12}{13}.

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β=45⋅1213−35⋅513=48−1565=3365\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\tfrac45\cdot\tfrac{12}{13}-\tfrac35\cdot\tfrac5{13}=\tfrac{48-15}{65}=\tfrac{33}{65} → D.

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