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IDENTITAS TRIGONOMETRI

IDENTITAS TRIGONOMETRI

Assessment

Presentation

•

Mathematics

•

10th - 12th Grade

•

Hard

Created by

Intan Ryanti

Used 8+ times

FREE Resource

6 Slides • 20 Questions

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IDENTITAS TRIGONOMETRI

by Intan Ryanti

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​PENJUMLAHAN DAN SELISIH DUA SUDUT
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Multiple Choice

sin⁡ (A−B)=\sin\ \left(A-B\right)=  

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sin⁡ A cos⁡ A−sin⁡ B cos⁡ B\sin\ A\ \cos\ A-\sin\ B\ \cos\ B  

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sin⁡ A cos⁡ B + sin⁡ A cos⁡ B\sin\ A\ \cos\ B\ +\ \sin\ A\ \cos\ B  

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sin⁡ A cos⁡ B − cos⁡ A sin⁡ B\sin\ A\ \cos\ B\ -\ \cos\ A\ \sin\ B  

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sin⁡ A cos⁡ B +cos⁡ A sin⁡ B\sin\ A\ \cos\ B\ +\cos\ A\ \sin\ B  

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Multiple Choice

cos⁡ (A−B)=\cos\ \left(A-B\right)=  

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cos⁡ A cos⁡ B −sin⁡ A sin⁡ B\cos\ A\ \cos\ B\ -\sin\ A\ \sin\ B  

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cos⁡ A cos⁡ B +sin⁡ A sin⁡ B\cos\ A\ \cos\ B\ +\sin\ A\ \sin\ B  

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cos⁡ A sin⁡ B −sin⁡ A cos⁡ B\cos\ A\ \sin\ B\ -\sin\ A\ \cos\ B  

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sin⁡ a sin⁡ B −cos⁡ a cos⁡ B\sin\ a\ \sin\ B\ -\cos\ a\ \cos\ B  

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Multiple Choice

tan⁡ (A+B)=\tan\ \left(A+B\right)=  

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tan⁡ A +tan⁡ B1−tan⁡Atan⁡ B\frac{\tan\ A\ +\tan\ B}{1-\tan A\tan\ B}  

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tan⁡ A−tan⁡ B1−tan⁡Atan⁡B\frac{\tan\ A-\tan\ B}{1-\tan A\tan B}  

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tan⁡ A+tan⁡B1+tan⁡Atan⁡B\frac{\tan\ A+\tan B}{1+\tan A\tan B}  

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tan⁡ A−tan⁡B1+tan⁡ Atan⁡ B\frac{\tan\ A-\tan B}{1+\tan\ A\tan\ B}  

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Multiple Choice

Uraikan bentuk berikut! cos⁡ (2x+2y)=\cos\ \left(2x+2y\right)=  

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cos⁡ 2x cos⁡ 2y+sin⁡ 2x sin⁡ 2y\cos\ 2x\ \cos\ 2y+\sin\ 2x\ \sin\ 2y  

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cos⁡ 2x cos⁡ 2y−sin⁡ 2x sin⁡ 2y\cos\ 2x\ \cos\ 2y-\sin\ 2x\ \sin\ 2y  

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sin⁡ 2x cos⁡ 2y−cos⁡ 2x sin⁡ 2y\sin\ 2x\ \cos\ 2y-\cos\ 2x\ \sin\ 2y  

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sin⁡ 2x cos⁡ 2y +cos⁡ 2x sin⁡ 2y\sin\ 2x\ \cos\ 2y\ +\cos\ 2x\ \sin\ 2y  

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Multiple Choice

sin⁡ (x+2y)=\sin\ \left(x+2y\right)=  

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sin⁡ x cos⁡ 2y −cos⁡ x sin⁡ 2y\sin\ x\ \cos\ 2y\ -\cos\ x\ \sin\ 2y  

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sin⁡ x cos⁡ 2y +cos⁡ x sin⁡ 2y\sin\ x\ \cos\ 2y\ +\cos\ x\ \sin\ 2y  

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sin⁡ x sin⁡ 2y −cos⁡ x cos⁡ 2y\sin\ x\ \sin\ 2y\ -\cos\ x\ \cos\ 2y  

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cos⁡ x cos⁡ 2y +sin⁡ x sin⁡ 2y\cos\ x\ \cos\ 2y\ +\sin\ x\ \sin\ 2y  

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Multiple Choice

tan⁡ (2x−y)=\tan\ \left(2x-y\right)=  

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tan⁡ 2x−tan⁡ y1−tan⁡2x tan⁡ y\frac{\tan\ 2x-\tan\ y}{1-\tan2x\ \tan\ y}  

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tan⁡ 2x +tan⁡y1+tan⁡ 2x tan⁡y\frac{\tan\ 2x\ +\tan y}{1+\tan\ 2x\ \tan y}  

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tan⁡ 2x −tan⁡ y1+tan⁡ 2x tan⁡ y\frac{\tan\ 2x\ -\tan\ y}{1+\tan\ 2x\ \tan\ y}  

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tan⁡ 2x +tan⁡ y1−tan⁡ 2x tan⁡ y\frac{\tan\ 2x\ +\tan\ y}{1-\tan\ 2x\ \tan\ y}  

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Multiple Choice

sin⁡ 15°=\sin\ 15\degree=  

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sin⁡ (45−30)=sin⁡ 45 cos⁡ 30−cos⁡ 45 sin⁡ 30\sin\ \left(45-30\right)=\sin\ 45\ \cos\ 30-\cos\ 45\ \sin\ 30  

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sin⁡ (45+30)=sin⁡ 45 cos⁡ 30+cos⁡ 45 sin⁡ 30\sin\ \left(45+30\right)=\sin\ 45\ \cos\ 30+\cos\ 45\ \sin\ 30  

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sin⁡ (45−30)=sin⁡ 45 sin⁡ 30−cos⁡ 45 cos⁡ 30\sin\ \left(45-30\right)=\sin\ 45\ \sin\ 30-\cos\ 45\ \cos\ 30  

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sin⁡ (45−30)=cos⁡ 45 cos⁡ 30−sin⁡ 45 sin⁡ 30\sin\ \left(45-30\right)=\cos\ 45\ \cos\ 30-\sin\ 45\ \sin\ 30  

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Multiple Choice

cos⁡ 75°=\cos\ 75\degree=  

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14(6−2)\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)  

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12(6−2)\frac{1}{2}\left(\sqrt{6}-\sqrt{2}\right)  

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14(6+2)\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)  

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14(3−2)\frac{1}{4}\left(\sqrt{3}-\sqrt{2}\right)  

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Multiple Choice

cos⁡85°cos⁡55° + sin⁡85°sin⁡55° = . . .\cos85\degree\cos55\degree\ +\ \sin85\degree\sin55\degree\ =\ .\ .\ .  

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00  

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11  

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12\frac{1}{2}  

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122\frac{1}{2}\sqrt{2}  

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123\frac{1}{2}\sqrt{3}  

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Multiple Choice

tan⁡75°−tan⁡15°1+tan⁡75°tan⁡15°= . . .\frac{\tan75\degree-\tan15\degree}{1+\tan75\degree\tan15\degree}=\ .\ .\ .  

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00  

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3\sqrt{3}  

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133\frac{1}{3}\sqrt{3}  

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11  

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−3-\sqrt{3}  

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​PERKALIAN SINUS DAN COSINUS
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​CONTOH
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Multiple Choice

2sin⁡xcos⁡3x=.....2\sin x\cos3x=.....  

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cos⁡4x−cos⁡2x\cos4x-\cos2x  

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cos⁡4x+cos⁡2x\cos4x+\cos2x  

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sin⁡4x+sin⁡2x\sin4x+\sin2x  

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sin⁡4x−sin⁡2x\sin4x-\sin2x  

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sin⁡2x−sin⁡4x\sin2x-\sin4x  

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Multiple Choice

Bentuk lain dari  2 sin3a sin a  adalah .....

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cos 4a - cos 2a

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cos 2a - cos 4a

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sin 4a - sin 2a

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sin 4a + sin 2a

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cos 4a + cos 2a

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Multiple Choice

sin⁡ 75° sin⁡ 15°= ....\sin\ 75\degree\ \sin\ 15\degree=\ ....  

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34\frac{3}{4}  

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12\frac{1}{2}  

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14\frac{1}{4}  

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−14-\frac{1}{4}  

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−34-\frac{3}{4}  

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Multiple Choice

cos⁡ 75° sin⁡ 15°= ....\cos\ 75\degree\ \sin\ 15\degree=\ ....  

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2+34\frac{2+\sqrt{3}}{4}  

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2−34\frac{2-\sqrt{3}}{4}  

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3−24\frac{3-\sqrt{2}}{4}  

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3+34\frac{3+\sqrt{3}}{4}  

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3−34\frac{3-\sqrt{3}}{4}  

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Multiple Choice

nilai dari  2 sin⁡ 50o cos⁡ 40o + 2 cos⁡ 95o sin⁡ 85o = .....nilai\ dari\ \ 2\ \sin\ 50^o\ \cos\ 40^o\ +\ 2\ \cos\ 95^o\ \sin\ 85^o\ =\ .....

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−1-1  

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−12-\frac{1}{2}  

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12\frac{1}{2}  

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11  

5

22  

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​PENJUMLAHAN/SELISISH SINUS DAN COSINUS
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Multiple Choice

Nyatakan 2cos45°.cos15° sebagai bentuk penjumlahan, kemudian tentukan nilainya!

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cos60° + cos 30° dan ½ (1 + √3)

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cos60° + cos 30° dan (1 + √3)

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Multiple Choice

Nyatakan 2sin80°. cos50° sebagai bentuk penjumlahan!

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sin130° + sin 30°

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sin100° + sin 30°

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Multiple Choice

Sederhanakanlah sin155° - sin25° !

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0

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sin 65°

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Multiple Choice

Rumus sinus jumlah dua sudut...

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sin⁡ (α+β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β\sin\ \left(\alpha+\beta\right)=\sin\ \alpha\ \cos\ \beta+\cos\ \alpha\ \sin\ \beta

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sin⁡ (α−β)=sin⁡α cos⁡ β−cos⁡ α sin⁡ β\sin\ \left(\alpha-\beta\right)=\sin\alpha\ \cos\ \beta-\cos\ \alpha\ \sin\ \beta

3

sin⁡ (α+β)=sin⁡ α cos⁡ β−cos⁡ α sin⁡ β\sin\ \left(\alpha+\beta\right)=\sin\ \alpha\ \cos\ \beta-\cos\ \alpha\ \sin\ \beta

4

sin⁡ (α−β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β\sin\ \left(\alpha-\beta\right)=\sin\ \alpha\ \cos\ \beta+\cos\ \alpha\ \sin\ \beta

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Multiple Choice

Rumus tangen jumlah dua sudut...

1

tan⁡ (α+β)=tan⁡ α−tan⁡ β1+tan⁡ α tan⁡ β\tan\ \left(\alpha+\beta\right)=\frac{\tan\ \alpha-\tan\ \beta}{1+\tan\ \alpha\ \tan\ \beta}

2

tan⁡ (α+β)=tan⁡β+tan⁡ α 1+tan⁡ α tan⁡ β\tan\ \left(\alpha+\beta\right)=\frac{\tan\beta+\tan\ \alpha\ }{1+\tan\ \alpha\ \tan\ \beta}

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tan⁡ (α+β)=tan⁡ α+tan⁡ β1−tan⁡ α tan⁡ β\tan\ \left(\alpha+\beta\right)=\frac{\tan\ \alpha+\tan\ \beta}{1-\tan\ \alpha\ \tan\ \beta}

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tan⁡ (α−β)=tan⁡ α−tan⁡ β1+tan⁡ α tan⁡ β\tan\ \left(\alpha-\beta\right)=\frac{\tan\ \alpha-\tan\ \beta}{1+\tan\ \alpha\ \tan\ \beta}

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IDENTITAS TRIGONOMETRI

by Intan Ryanti

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