Jawaban Uji Kompetensi 8 3

Oleh Lilik Nur Fadlilah

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Transkrip Jawaban Uji Kompetensi 8 3

JAWABAN UJI KOMPETENSI 8.3 1. SINUS, COSINUS, TANGEN, SECAN, COSEC, COTANGEN a. y r = √ 52 +62 A (5,6) 6 = √ 25+36 r=? =√ 61 0 5 x y b. P (-3,8) r = √ (−3)2 +82 = √ 9+64 r=? 8 = √ 73 -3 0 x c. y r = √ (−3)2 +(−5)2 x (-5,-3) r = ? =√ 34 -5 T =√ 9+25 -3 d. x r = √ 22+(−8)2 r=? = √ 4 +64 = √ 68 -8 y 2 Q (2,-8) JAWAB: 6 √61 6 √ 61 x = √61 √61 61 5 √61 5 √ 61 Cos A = x = √61 √61 61 6 Tan A = 5 61 Sec A = √ 5 a) Sin A = 61 Cosec A = √ 6 5 Cot A = 6 8 √73 8 √ 73 x = √73 √73 73 −3 √73 −3 √ 73 2 Cos P = x = √ 73 √73 73 8 −8 3 Tan P = −3 = 3 73 4 Cosec P = √ 8 −√ 73 5 Sec P = 3 −3 6 Cot P = 8 b) 1 Sin P = −5 √34 −5 √ 34 x = √ 34 √34 34 −3 √34 −3 √ 34 Cot T = x = √34 √34 34 −5 5 Tan T = −3 = 3 34 − 34 Sec T = √ = √ −3 3 34 − 34 Cosec T = √ = √ −5 5 −3 3 Cot T = −5 = 5 c) Sin T = d) Sin Q = 2 √68 2 √ 68 2 √ 68 x = = 34 √68 √68 68 −8 √ 68 −8 √ 68 −2 x = = 17 √68 √ 68 68 2 1 Tan Q = −8 =- 4 68 − 68 Sec Q = √ = √ −8 8 68 Cosec Q = √ 2 −2 Cot Q = 2 = -4 Cos Q = √ 68 2. A. P (5,12) y 12 P (5,12) r = √ 52 +122 = √ 25+144 12 = √ 164 = 13 5 Sin P = 12 13 Cos P = 5 13 Tan P = x 12 5 B. Q (-5.2,7.2) r = √ (7,2)2+(−5,2)2 Q (-5.2,7.2) 7,2 7,2 = √ 51,84+27,04 = √ 78,88 = √ 79 -5,2 Sin Q = 7,2 √79 7,2 √ 79 x = √79 √79 79 Cos Q = −5,2 √79 −5,2 x = 79 √ 79 √ 79 √79 −7,2 Tan Q = 5,2 C. R (-5,-2) 5√ 29 (-5,-2) R -2 -5 Sin R = −2 √ 29 −2 x = √ 29 √29 √ 29 29 Cos R = −5 √ 29 −5 √ 29 x = √29 √29 29 Tan R = −2 2 = −5 5 D. T (3.5,-7.75) 3,5 r = √ (3,5)2 +(−7,75)2 r=? = √ 12,25+60,0625 = √ 72,3125 -7,75 Sin T = Cos T = −7,75 √ 72 −7,75 x = 72 √ 72 √ 72 √72 3 ,5 √ 72 3 ,5 √ 72 x = √72 √ 72 72 T = √ 72 −7,75 Tan T = 3,5 3. a. Karena pada kuadran IV nilai tanda untuk Sin x dan Cos x bertanda negatif (-), 1 yang mana Sec = cos x b. Tidak karena antara nilai Sinus dan Cosinus nilainya berkebalikan c. 4. Sin α > 0 Cos α >0       Sin α < 0 Cos α < 0 Tanα < 0 Sinα > 0 Sin α > 0 → 00 < α < 900 & 900 < α < 1800 Cos α > 0 → 00 < α < 900 & 2700 < α < 3600 Sin α < 0 → 1800 < α < 3600 Cos α < 0 → 900 < α < 2700 Tan α < 0 → 900 < α < 1800 & 2700 < α < 3600 Sin α > 0 → 00 < α < 900 & 900 < α < 1800 −8 5. Tan x = 15 Sin α > 0 sin α 8 r = √ 64+225 y = cos x = 15 = x = √ 289 = 17 −15  Cos α = 17 17  Sec α = −15 17  Cosec α = 8 17 8 Cosec α 17  Cotan α = −15 = −15 8 8 −15  ( sin α ) (cos α) = 17 ∙ 17 120 = 289 = 0,4 Kuadran II & I Kuadran IV & I Kuadran III & IV Kuadran II & III Kuadran II & IV Kuadran II & 1 −15  Cotan α = 8 π π ≤ β ≤ 3 ; nilai Cotan tidak terdefinisi 2 2 0 Misal : β = 180 1 Ambil , Cotan β = tan β 1 = tan 180 ° 1 = 0 → tak terdefinisi a. Sin β = Sin 1800 6. Diket : =1 b. Cos β = Cos 1800 =0 sin β sin 180 ° 1 c. tan β +1 = tan(80° +1) = 0+1 = 1 2 cos β tan β−1 2 = cos β x tan β−1 2(tan β−1) = cos β 2(tan 180 °−1) = cos 180 ° 2(0−1) = 0 2(0−1) −2 = = 0 → tak terdefinisi 0 7. A. Cos x ∙ Cosec x ∙ Tan x 1 sin x = Cos x ∙ sin x ∙ cos x cos x sin x 1 = sin x ∙ cos x = 1 = 1 B. Cos x ∙ Cotan x ∙ Sin x 1 = Cos x ∙ tan x + Sin x 1 = Cos x ∙ sin x + Sin x cos x 2 Sec β d. tan β−1 = 2 cos x = + Sin x sin x 2 cos x+ sin x = sin x 1 = sin x = Cosec x − 3 −x 8. Cos β = √ = r 4 x = -√ 3 r =4 y = √ 4 2−¿ ¿ )2 = √ 16−3 =√ 13 √  Sec β = cos β = x = 3 √3 √ 3 √3 −4 1 3 −4 − 3  Sin β = √ 4 −√ 3 √ 3 √3 1 √ 3 9  Tan β = = x = −√ 3 √ 3 √3 3  Cos 15 = 60-45 1 1 = 2 - 2 √2 45-30 1 1 1 = 2 - 2 ∙ √2 = √2 a. 2 Sec β−tan β + Sec β tan β −4 4 −¿ ¿ + (- ) √3 √3 −4 13 4 −12−13 √3−12 − − 3 3 √3 √3 √3 = √ 13 √13 √3 √3 −24−13 √ 3 √ 3 = x 3 √3 √3 = ¿¿ = −24 √ 3−39 √ 39 x 3 √39 √ 39 −24 √ 117−39 √ 39 = 117 = 1 = 2 √2 - 2 √3 2 2 Sec β + Sec β−1 Sec 2 β+ tan 2 β b. = 2 2 2 2sin β +2 cos β 2(sin ¿ ¿ 2 β+ cos β) ¿ = 2 Sec 2 β−1 2−1 2 −4 3) −1 2 Sec 2 β−1 2( √ = = 3 2−1 2 2 −4 2( 3) −1 √ = 3 2 ❑ −16 2( 3) −1 9 = 2 ❑ −16 2( 2.1) −1 3 = 2 −32 3 − 35 1 35 = 3 3 =-3 x 2=- 6 2 9. Jika α = 2040 , Hitunglah: −1 √3 −1 √ 3 sin α sin 240 2 2 a. = -2√ 3 2 = 2 = 2 = 1 (cos α ) (cos 240) −1 ( ) 4 2 1 1 + √3 2 2 ❑ ❑ ❑ α = 240 = −1 1 x 1 1 ❑ cos α + sin( ) cos 240+sin( ) + √3 4 4 + √3 2 2 2 2 tan α b. √3 tan 240 1 1 1 3 √3+ ∙ 3 √3+ 2 2 2 2 = 1 1 ❑ = 1 3❑ − + ∙3 − + 4 4 4 4 1 3 1 3 √ 3+ √3+ 2 2 2 2 = = 2 1 4 2 1 3 = 2 (2 √3 + 2) = √3 + 3 c. 2 sin α - cos (2 α) = 2 sin 240 - cos 480 1 1 = 2 ∙ - 2 √ 3 - (- 2 ¿ 1 = - √3 + 2 d. Sin2 α + Cos2 α + √ 3 = Sin2 240 + Cos2 240 + √ 3 −1 −1 2 2 = ( 2 √ 3¿ ¿ + ( 2 ¿ + √ 3 1 1 = 4 ∙ 3 +4 + √3 3 1 = 4 + 4 +√ 3 4 = 4 + √3 = 1 + √3 sin A sin A 10. a. 1+ cos A + 1−cos A = = = sin A ( 1−cos A )+ sin A (1 cos A) ( 1+cos A ) (1−cos A ) sin A ( 1−cos A )+ sin A (1 cos A) 2 1−cos A +cos A−C 0 s A sin A ( 1−cos A )+ sin A (1 cos A) 2 1−C 0 s A sin A – sin A cos A+sin A+sin A cos A 2 sin A = 2 2 1−C 0 s A 1−C 0 s A 2 sin A = 2 2 2 C 0 s A+ sin A−C 0 s A 2sin A = 2 sin A 2 = sin A = = 2 Cosec A b . (Sin B + Cos B)2 + (Sin B – Cos B)2 = Sin2 B + 2 Sin B Cos B + Cos2 B + Sin2 A – 2 Sin B Coc B + Cos2 B = 2 Sin2 B + 2 Cos2 B = 2 (Sin2 B + Cos2 B) = 2 (1) =2 c. (Cosec A – Cotan A) (1 + Cos A) = Cosec A + Cosec A Cos A – Cotan A Cos A 1 1 1 1 cos A cos A 1 cos A cos2 A = sin A + sin A Cos A - tan A ∙ Cos A = sin A + sin A - sin A ∙ Cos A = sin A + sin A sin A = sin2 A+ cos2 A +cos A−cos A sin A 2 2 cos A sin A+ cos A sin A = = + sin A = Sin A + Cotan A sin A sin A 11. Dggnmj,k 12.y = 3 Cos 2x – 2 a). y maks. Dicapai bila Cos x = 1 Cos 2x = 2 Cos2 x – 1 = 2 ∙ 1-1 =1 y min. Dicapai bila Cos x = -1 Cos 2x = 2 Cos2x – 1 = 2(-1)2 – 1 = 2-1 =1 b). y = 5 Sin x + Cos 2x y maks. dicapai bila Sin x = 1 , y maks = 3∙ 1 - 2 = 3-2 =1 y min= 3 Cos 2x-2 = 3∙ 1 - 2 = 3-2 =1 Cos x = 1 → Cos 2x = 2 Cos2 x-1 = 2 ∙ 12 – 1 = 1 y maks. = 5 ∙ 1 + 1 = 6 y min dicapai bila Sin x = - 1 Cos x = - 1 → Cos 2x = 2 Cos2 x – 1 = 2 (-1)2 – 1 =1 c). 7 Sinx+cos x d).y = Sinx−cos x x Sinx+cos x = 7¿¿ = 7(Sinx+Cosx) 7( Sinx+Cosx) = 2 sin x +sin 2 x−1 2sin 2 x−1 y maks. dicapai bila Sin x = 1 , y maks = 7.2 7(−1−1) = 2−1 = 14 2.1−1 y min dicapai bila Sin x = -1 , y min = Cos x = 1 7(−1−1) = 7(-2) 2.1−1 Cos x = -1 = -14 13.a. √ 2 Sin 2x = tg 2x sin 2 x √ 2 Sin 2x = cos 2 x √ 2 cos 2 x = 1 1 Cos 2x = 2 √ 2 2x = 45° x = 22,5° b. (Cos x + Sin x )2 = 12 Cos2x + 2 Cos x Sin x + Sin2x = 1 2 Cos x Sin x + 1 = 1 2 Cos x Sin x =0 Sin 2x = 0 2x = 0° , 180° , 360° , 540° , 720° x = 0° , 90° , 180° , 270° , 360° 2 2 c Sin x + Cos x = 1 Cos 2x – Cos2 x + Cos2x = 1 Cos 2x = 1 2x = 0° , 360° x = 0° , 180°

Judul: Jawaban Uji Kompetensi 8 3

Oleh: Lilik Nur Fadlilah


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