Pertanyaan

Tentukan bentuk sederhana dari: sec 13 5 ∘ + tan 15 0 ∘ sin 15 0 ∘ + cos 12 0 ∘

Tentukan bentuk sederhana dari:

$\frac{sin 15 0 ^{\circ} + cos 12 0 ^{\circ}}{sec 13 5 ^{\circ} + tan 15 0 ^{\circ}}$

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S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

.

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator size 14px sin size 14px space size 14px 150 size 14px degree size 14px plus size 14px cos size 14px space size 14px 120 size 14px degree over denominator size 14px sec size 14px space size 14px 135 size 14px degree size 14px plus size 14px tan size 14px space size 14px 150 size 14px degree end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank size 14px equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank size 14px 0 end table$ .

Pembahasan

Ingatlah rumus sudut berelasi . Karena , maka dan Lalu karena , maka Kemudian, karena , maka Akibatnya Jadi, .

Ingatlah rumus sudut berelasi

Karena , maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 150 degree end cell equals cell sin space left parenthesis 180 degree minus 30 degree right parenthesis end cell row blank equals cell sin space 30 degree end cell row blank equals cell 1 half end cell end table end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 150 degree end cell equals cell tan space left parenthesis 180 degree minus 30 degree right parenthesis end cell row blank equals cell negative tan space 30 degree end cell row blank equals cell negative 1 third square root of 3 end cell end table end style

Lalu karena , maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 120 degree end cell equals cell cos space left parenthesis 180 degree minus 60 degree right parenthesis end cell row blank equals cell negative cos space 60 degree end cell row blank equals cell negative 1 half end cell end table end style

Kemudian, karena , maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sec space 135 degree end cell equals cell sec space left parenthesis 180 degree minus 45 degree right parenthesis end cell row blank equals cell negative sec space 45 degree end cell row blank equals cell negative square root of 2 end cell end table end style

Akibatnya

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 150 degree plus cos space 120 degree over denominator sec space 135 degree plus tan space 150 degree end fraction end cell equals cell fraction numerator begin display style 1 half end style plus open parentheses negative begin display style 1 half end style close parentheses over denominator open parentheses negative square root of 2 close parentheses plus open parentheses negative begin display style 1 third end style square root of 3 close parentheses end fraction end cell row blank equals 0 end table end style$

Jadi, $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator size 14px sin size 14px space size 14px 150 size 14px degree size 14px plus size 14px cos size 14px space size 14px 120 size 14px degree over denominator size 14px sec size 14px space size 14px 135 size 14px degree size 14px plus size 14px tan size 14px space size 14px 150 size 14px degree end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank size 14px equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank size 14px 0 end table$ .

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