Pertanyaan

Diketahui dengan 0° ≤ x ≤ 90°. Nilai dari cos 2 x = ....

Diketahui dengan 0° ≤ x ≤ 90°. Nilai dari $cos \frac{x}{2} =$ ....

$begin mathsize 14px style negative square root of fraction numerator 7 plus 2 square root of 7 over denominator 14 end fraction end root end style$
$begin mathsize 14px style negative square root of fraction numerator 2 plus 7 square root of 2 over denominator 14 end fraction end root end style$
$begin mathsize 14px style square root of fraction numerator 2 plus 2 square root of 7 over denominator 14 end fraction end root end style$
$begin mathsize 14px style square root of fraction numerator 2 plus 7 square root of 2 over denominator 14 end fraction end root end style$
$begin mathsize 14px style square root of fraction numerator 7 plus 2 square root of 7 over denominator 14 end fraction end root end style$

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Klaim

D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Perhatikan bahwa Sehingga dibutuhkan nilai dari cos x terlebih dahulu. Karena 0° ≤ x ≤ 90° , maka x adalah sudut lancip. Perhatikan segitiga berikut Karena ,maka Karena segitiga di atas merupakan segitiga siku-siku, maka nilai y bisa didapat menggunakan teorema Pythagoras. Sehingga Sehingga didapatkan . Maka Karena 0° < x < 90°, maka Sehingga merupakan sudut pada kuadran I. Maka cos⁡ bernilai positif. Sehingga

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses x plus 30 degree close parentheses end cell equals cell 1 half sin invisible function application x end cell row cell cos invisible function application x cos invisible function application 30 degree minus sin invisible function application x sin invisible function application 30 degree end cell equals cell 1 half sin invisible function application x end cell row cell cos invisible function application x times 1 half square root of 3 minus sin invisible function application x times 1 half end cell equals cell 1 half sin invisible function application x end cell row cell 1 half square root of 3 cos invisible function application x minus 1 half sin invisible function application x end cell equals cell 1 half sin invisible function application x end cell row cell 1 half square root of 3 cos invisible function application x end cell equals cell 1 half sin invisible function application x plus 1 half sin invisible function application x end cell row cell 1 half square root of 3 cos invisible function application x end cell equals cell sin invisible function application x end cell row cell 1 half square root of 3 end cell equals cell fraction numerator sin invisible function application x over denominator cos invisible function application x end fraction end cell row cell tan invisible function application x end cell equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table end style$

Perhatikan bahwa

$begin mathsize 14px style cos invisible function application x over 2 equals plus-or-minus square root of fraction numerator 1 plus cos invisible function application x over denominator 2 end fraction end root end style$

Sehingga dibutuhkan nilai dari cos x terlebih dahulu.

Karena 0° ≤ x ≤ 90°, maka x adalah sudut lancip. Perhatikan segitiga berikut

Karena $begin mathsize 14px style tan invisible function application x equals fraction numerator square root of 3 over denominator 2 end fraction end style$ , maka

Karena segitiga di atas merupakan segitiga siku-siku, maka nilai y bisa didapat menggunakan teorema Pythagoras. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of 2 squared plus open parentheses square root of 3 close parentheses squared end root end cell row blank equals cell square root of 4 plus 3 end root end cell row blank equals cell square root of 7 end cell end table end style

Sehingga didapatkan $begin mathsize 14px style cos invisible function application x equals 2 over y equals fraction numerator 2 over denominator square root of 7 end fraction end style$ .

Maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application x over 2 end cell equals cell plus-or-minus square root of fraction numerator 1 plus cos invisible function application x over denominator 2 end fraction end root end cell row blank equals cell plus-or-minus square root of fraction numerator 1 plus fraction numerator 2 over denominator square root of 7 end fraction over denominator 2 end fraction end root end cell row blank equals cell plus-or-minus square root of fraction numerator 1 plus fraction numerator 2 over denominator square root of 7 end fraction times fraction numerator square root of 7 over denominator square root of 7 end fraction over denominator 2 end fraction end root end cell row blank equals cell plus-or-minus square root of fraction numerator 1 plus 2 over 7 square root of 7 over denominator 2 end fraction end root end cell row blank equals cell plus-or-minus square root of fraction numerator 1 plus 2 over 7 square root of 7 over denominator 2 end fraction times 7 over 7 end root end cell row blank equals cell plus-or-minus square root of fraction numerator 7 plus 2 square root of 7 over denominator 14 end fraction end root end cell end table end style$

Karena 0° < x < 90°, maka

$begin mathsize 14px style 0 degree less than x less than 90 degree fraction numerator 0 degree over denominator 2 end fraction less than x over 2 less than fraction numerator 90 degree over denominator 2 end fraction 0 degree less than x over 2 less than 45 degree end style$

Sehingga merupakan sudut pada kuadran I. Maka cos⁡ bernilai positif. Sehingga

$begin mathsize 14px style cos invisible function application x over 2 equals square root of fraction numerator 7 plus 2 square root of 7 over denominator 14 end fraction end root end style$