rumus fungsi turunan terigonometri dari f(x)=5 sin x?
1. rumus fungsi turunan terigonometri dari f(x)=5 sin x?
Penjelasan dengan langkah-langkah:
f(x) = 5sin(x)
f'(x) = 1.5cos(x)
f'(x) = 5cos(x)
2. rumus fungsi turunan terigonometri dari Y=2x+cos x dan h(x)=cos ²x?
Jawaban:
cos 30 = cos 30 1/2 akar 3
cos 45 = 1/2 akar 2
cos 60 = cos 60 = 1/2
sin 90 = sin 90 =1
Penjelasan dengan langkah-langkah:
semoga membantu maaf kalau salah
3. tentuka perbandingan terigonometri di sudut a
a)
[tex] \sin( \alpha ) = \frac{12}{13} [/tex]
b)
[tex] \cos( \alpha ) = \frac{5}{13} [/tex]
c)
[tex] \tan( \alpha ) = \frac{12}{5} [/tex]
d)
[tex] \csc( \alpha ) = \frac{1}{ \sin( \alpha ) } = \frac{1}{ \frac{12}{13} } = \frac{1 \times 13}{12} = \frac{13}{12} [/tex]
e)
[tex] \sec( \alpha) = \frac{1}{ \cos( \alpha ) } = \frac{1}{ \frac{5}{13} } = \frac{1 \times 13}{5} = \frac{13}{5} [/tex]
f)
[tex] \cot( \alpha ) = \frac{1}{ \tan( \alpha ) } = \frac{1}{ \frac{12}{5} } = \frac{1 \times 5}{12} = \frac{5}{12} [/tex]
4. gambarlah tabel dan grafik terigonometri dari f (x)=2 cos x + 1
Semoga membantu kak. Kalau kurang jelas boleh ditanyakan
5. Tentukan nilai bentuk terigonometri berikut :
Jawaban:
Jawaban terlampir, jadikan jawaban terbaik yaa
6. Tentukan nilai bentuk terigonometri berikut 12 sin 15° sin 165°
Penjelasan dengan langkah-langkah:
[tex]-2 sin (a) sin (b) = cos ( a - b ) - cos ( a + b )[/tex]
[tex] \frac{1}{2} \sin(a) \sin(b) = \cos(a - b) - \cos(a + b) [/tex]
Jawab :[tex] = 12 \times \frac {1}{2} \times (\cos(15- 165) - \cos(15 + 165) )[/tex]
[tex] = 6 \: ( \cos( - 150) - \cos(180) [/tex]
[tex] = 6 \: ( \frac{ \sqrt{3} }{2} - ( - 1))[/tex]
[tex] = 6 \: ( \frac{ \sqrt{3} }{2} + 1)[/tex]
[tex] = - 3 \sqrt{3} + 6[/tex]
7. igonometri sin x cosec x - sin² x = cos² x
sinx . cscx - sin²x = cos²x
sin ( 1/sinx) - sin²x = cos²x
sinx/sinx - sin²x = cos²x
1 - sin²x = cos²x
cos²x = cos²x
8. Tentukan nilai bentuk terigonometri berikut :
Jawaban:
[tex] \sf \: \frac{ \sin(195) - \sin(45) }{ \cos(285) - \cos(135) } \\ \\ \sf \: \frac{ \sin(195) - \frac{ \sqrt{2} }{2} }{ \cos(285) - ( - \frac{ \sqrt{2} }{2} )} \\ \\ \sf \: \frac{ \frac{2 \sin(195) - \sqrt{2} }{2} }{ \cos(285) + \frac{ \sqrt{2} }{2} } \\ \\ \sf \: \frac{ \frac{2 \sin(195) - \sqrt{2} }{2} }{ \frac{2 \cos(285) + \sqrt{2} }{2} } \\ \\ \sf \: \frac{2 \sin(195) - \sqrt{2} }{2 \cos(285) + \sqrt{2} } \\ \\ \sf \: \frac{2 \sin(45 + 150) - \sqrt{2} }{2 \cos(45 + 240) + \sqrt{2} } \\ \\ \sf \: \frac{2( \sin(45) \cos(150) + \cos(45) + \sin(150)) - \sqrt{2} }{2 (\cos(45) \cos(240) - \sin(45) \sin(240) ) + \sqrt{2} } \\ \\ \sf \: \frac{2( \frac{ \sqrt{2} }{2}( - \frac{ \sqrt{3} }{2}) + \frac{ \sqrt{2} }{2} \times \frac{1}{2} ) - \sqrt{2} }{2( \frac{ \sqrt{2} }{2}( - \frac{1}{2} ) - \frac{ \sqrt{2} }{2} \times ( - \frac{ \sqrt{3} }{2} )) + \sqrt{2} } \\ \\ \sf \: \frac{2( - \frac{ \sqrt{6} }{4} + \frac{ \sqrt{2} }{4} ) - \sqrt{2} }{2( - \frac{ \sqrt{2} }{4} + \frac{ \sqrt{6} }{4} ) + \sqrt{2}} = \frac{2( \frac{ - \sqrt{6} + \sqrt{2} }{4} ) - \sqrt{2} }{2( \frac{ - \sqrt{2} + \sqrt{6} }{4} + \sqrt{2} } \\ \\ \sf \: \frac{ \frac{ - \sqrt{6} + \sqrt{2} }{2} - \sqrt{2} }{ \frac{ - \sqrt{2} + \sqrt{6} }{2} + \sqrt{2} } = \frac{ \frac{ - \sqrt{6} + \sqrt{2} - 2 \sqrt{2} }{2} }{ \frac{ - \sqrt{2} + \sqrt{6} + 2 \sqrt{2} }{2} } \\ \\ \sf \: \frac{ \frac{ - \sqrt{6} - \sqrt{2} }{2} }{ \frac{ \sqrt{2} + \sqrt{6} }{2} } = - 1[/tex]
9. Diketahui sin α = 15/18 dengan α sudut lancip. Hitunglah nilai terigonometri berikut!a. Sin (180°- α)b. Cos (380°-α)c. Sec (90°-α)d. Tan (180°+α)
Penjelasan dengan langkah-langkah:
[tex]x = \sqrt{ {18}^{2} - {15}^{2} } \\ = \sqrt{324 - 225 } \\ = \sqrt{99} = 3 \sqrt{11} [/tex]
a)
[tex] \sin(180 - \alpha ) = \sin( \alpha ) = \frac{15}{18} [/tex]
b)
[tex] \cos(360 - \alpha ) = \cos( \alpha ) = \frac{3 \sqrt{11} }{18} [/tex]
c)
[tex] \sec(90 - \alpha ) = \csc( \alpha ) \\ = \frac{1}{ \sin( \alpha ) } = \frac{18}{15} [/tex]
d)
[tex] \tan(180 + \alpha ) = \tan( \alpha ) \\ = \frac{15}{3 \sqrt{11} } \\ = \frac{15}{3 \sqrt{11} } \times \frac{ \sqrt{11} }{ \sqrt{11} } \\ = \frac{15}{33} \sqrt{11} = \frac{5}{11} \sqrt{11} [/tex]
