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Tentukan ekspresi aljabar dari setiap ekspresi berikut. a. cos ( tan − 1 ( − 3 2 x ​ ) )

Tentukan ekspresi aljabar dari setiap ekspresi berikut.

a.  

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W. Lestari

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Mahasiswa/Alumni Universitas Sriwijaya

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 cos space open parentheses tan to the power of negative 1 end exponent space open parentheses negative fraction numerator 2 x over denominator 3 end fraction close parentheses close parentheses equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 over denominator square root of 4 x squared plus 9 end root end fraction end cell end table.

Pembahasan

Ingat perbandingan sisi trigonometri berikut: Diketahui ekspresi . Misalkan: Perhatikan gambar berikut: Maka: Sehingga: Jadi, .

Ingat perbandingan sisi trigonometri berikut:

sin space theta equals fraction numerator sisi space depan over denominator sisi space miring end fraction rightwards arrow cosec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction cos space theta equals fraction numerator sisi space samping over denominator sisi space miring end fraction rightwards arrow sec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction tan space theta equals fraction numerator sisi space depan over denominator sisi space samping end fraction rightwards arrow cotan space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction 

Diketahui ekspresi cos space open parentheses tan to the power of negative 1 end exponent space open parentheses negative fraction numerator 2 x over denominator 3 end fraction close parentheses close parentheses. Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell tan to the power of negative 1 end exponent space open parentheses negative fraction numerator 2 x over denominator 3 end fraction close parentheses end cell row cell tan space theta end cell equals cell negative fraction numerator 2 x over denominator 3 end fraction space rightwards arrow fraction numerator sisi space depan over denominator sisi space samping end fraction end cell end table  

Perhatikan gambar berikut:

Maka:

p equals square root of open parentheses 2 x close parentheses squared plus 3 squared end root p equals square root of 4 x squared plus 9 end root 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses tan to the power of negative 1 end exponent space open parentheses negative fraction numerator 2 x over denominator 3 end fraction close parentheses close parentheses end cell equals cell cos space theta end cell row blank equals cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell row blank equals cell fraction numerator 3 over denominator square root of 4 x squared plus 9 end root end fraction end cell end table 

Jadi, cos space open parentheses tan to the power of negative 1 end exponent space open parentheses negative fraction numerator 2 x over denominator 3 end fraction close parentheses close parentheses equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 over denominator square root of 4 x squared plus 9 end root end fraction end cell end table.

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