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Identitas Trigonometri

Identitas Trigonometri

Assessment

Presentation

•

Mathematics

•

10th Grade

•

Practice Problem

•

Hard

Created by

Tsatsabilla Diannira

Used 6+ times

FREE Resource

16 Slides • 11 Questions

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Multiple Choice

Hasil dari sin⁡ x +tan⁡ xcot⁡ x+csc⁡ x=...\frac{\sin\ x\ +\tan\ x}{\cot\ x+\csc\ x}=...

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sin⁡ x. cos⁡ x\sin\ x.\ \cos\ x

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sin⁡ x. cot⁡ x\sin\ x.\ \cot\ x

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sin⁡ x. csc⁡ x\sin\ x.\ \csc\ x

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sin⁡ x. tan⁡ x\sin\ x.\ \tan\ x

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Multiple Choice

Hasil dari tan⁡α ⋅sin⁡αcos⁡ α ⋅sec⁡α=...\frac{\tan\alpha\ \cdot\sin\alpha}{\cos\ \alpha\ \cdot\sec\alpha}=...

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sin⁡2αcos⁡α\frac{\sin^2\alpha}{\cos\alpha}

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sin⁡2αcos⁡3α\frac{\sin^2\alpha}{\cos^3\alpha}

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sin⁡2α+sin⁡αcos⁡2α\frac{\sin^2\alpha+\sin\alpha}{\cos^2\alpha}

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sin⁡3α\sin^3\alpha

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Multiple Choice

Hasil dari 1−cos⁡ xsin⁡x=...Hasil\ dari\ \frac{1-\cos\ x}{\sin x}=...

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−sin⁡ x1+cos⁡x\frac{-\sin\ x}{1+\cos x}

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−cos⁡ x1−sin⁡x\frac{-\cos\ x}{1-\sin x}

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sin⁡ x1−cos⁡x\frac{\sin\ x}{1-\cos x}

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sin⁡ x1+cos⁡x\frac{\sin\ x}{1+\cos x}

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Multiple Choice

Bentuk (1−cos⁡2A)×cot⁡2A\left(1-\cos^2A\right)\times\cot^2A dapat disederhanakan menjadi ....

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2sin⁡2A−12\sin^2A-1

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sin⁡2A+cos⁡2A\sin^2A+\cos^2A

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1−sin⁡2A1-\sin^2A

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1−cos⁡2A1-\cos^2A

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Multiple Choice

Untuk setiap sudut β\beta , maka (1−sin⁡2β)(1+tan⁡2β)\left(1-\sin^2\beta\right)\left(1+\tan^2\beta\right) dapat disederhanakan menjadi ....

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1+sin⁡2β1+\sin^2\beta

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sin⁡2β−cos⁡2β\sin^2\beta-\cos^2\beta

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1+cos⁡2β1+\cos^2\beta

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Multiple Choice

tan⁡ x sin⁡ x+cos⁡ x\tan\ x\ \sin\ x+\cos\ x senilai dengan....

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cos⁡ x\cos\ x

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tan⁡ x\tan\ x

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sin⁡ x\sin\ x

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sec⁡ x\sec\ x

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Multiple Choice

sin⁡A1+cos⁡A+1+cos⁡Asin⁡A, hitunglah!\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A},\ hitunglah!

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2cos⁡A\frac{2}{\cos A}

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2cos⁡2A\frac{2}{\cos^2A}

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2sin⁡A\frac{2}{\sin A}

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2sin⁡2A\frac{2}{\sin^2A}

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Multiple Choice

bentuk yang senilai dengan 3cot⁡2α−2 adalahbentuk\ yang\ senilai\ dengan\ 3\cot^2\alpha-2\ adalah

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3sin⁡2α−5\frac{3}{\sin^2\alpha}-5

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3sin⁡2α−53\sin^2\alpha-5

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3sin⁡2α−23\sin^2\alpha-2

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3cos⁡2α−2\frac{3}{\cos^2\alpha}-2

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Multiple Choice

tan⁡2α cos⁡2α+cot⁡2α sin⁡2α, Hitunglah!\tan^2\alpha\ \cos^2\alpha+\cot^2\alpha\ \sin^2\alpha,\ Hitunglah!

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cos⁡2α+sin⁡2α\cos^2\alpha+\sin^2\alpha

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cos⁡2αsin⁡2α\frac{\cos^2\alpha}{\sin^2\alpha}

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1+tan⁡2α1+\tan^2\alpha

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Multiple Choice

Sederhanakan, cot⁡α−sec⁡2α1+cot⁡2aSederhanakan,\ \frac{\cot\alpha-\sec^2\alpha}{1+\cot^2a}

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sin⁡αcos⁡α\frac{\sin\alpha}{\cos\alpha}

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cos⁡αsin⁡α\frac{\cos\alpha}{\sin\alpha}

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1sin⁡α\frac{1}{\sin\alpha}

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sin⁡α\sin\alpha

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Multiple Choice

Diketahui sudut suatu bangun adalah β, maka (1+tan⁡2β)(1−sin⁡2β) dapat disederhanakan menjadiDiketahui\ sudut\ suatu\ bangun\ adalah\ \beta,\ maka\ \left(1+\tan^2\beta\right)\left(1-\sin^2\beta\right)\ dapat\ disederhanakan\ menjadi

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tan⁡2β\tan^2\beta

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cos⁡2β−1\cos^2\beta-1

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sin⁡2β+cos⁡2β\sin^2\beta+\cos^2\beta

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