Cosx=-√2/2

Cosx=-√2/2

(sinx+cosx)2+(sinx-cosx)2=2

1. (sinx+cosx)2+(sinx-cosx)2=2


Kelas 10 Matematika
Bab Trigonometri

(sin x + cos x)² + (sin x - cos x)²
= sin² x + 2 sin x cos x + cos² x + sin² x - 2 sin x cos x + cos² x
= (sin² x + cos² x) + (sin² x + cos² x) + 2 sin x cos x - 2 sin x cos x
= 1 + 1
= 2

TerbuktiPembuktian dapat dilihat pada lampiran

2. (sinx+cosx)^2 + (sinx-cosx)^2 = .....


(sinx+cosx)² + (sinx-cosx)² = .....
⇔ sin²x + 2sinx.cosx + cos²x + sin²x - 2sinx.cosx + cos²x
⇔ 2(sin²x + cos²x)
⇔ 2(1)
⇔ 2
(sin x + cos x) (sin x + cos x) + (sin x - cos x)(sin x - cos x)
= (sin^2 x + 2 sin x cos x + cos^2 x) + (sin^2x -2sin x cos x + cos ^2 x)
= sin ^2x + sin ^2x + cos ^2x +cos ^2x
= sin^2x + cos^2x + sin^2 x+ cos^2x
= 1+ 1
= 2

3. (cosx+sinx)^2 / (cosx-sinx)^2 = ⋯


jawab


(cos x  + sin x)^2 / (cos x - sin x)^2

= (1 + sin 2x)/ (1 - 2 sin  2x)

Trigonometri kelas 10

4. buktikan bahwa 2-(sinx + cosx)² = (sinx-cosx)²​


Jawab dan Penjelasan dengan langkah-langkah ada di gambar. Semoga membantu :)


5. (Sinx+cosx)2=1+2 sinx cosx


Sin^2 x + 2sinx.cosx +cos^2 x
Karena identitas triginometri
Sin^2 x +cos^2 x=1
Maka
Sin^2 x + 2sinx.cosx +cos^2
= 1 + 2sinx.cosx

6. Buktikan identitas trigonometri a. (cosx+sinx)(cosx-sinx)=1-2 sin^2x b. (cosx + sinx )^2 = 1+2cosx sinx


Identitas:
[tex]\sin^2 x+\cos^2 x=1[/tex]
Sehingga, berlaku juga (Jika perlu)
[tex]\sin^2 x=1-\cos^2x \\ \cos^2x=1-\sin^2x[/tex]
Maka,
Bagian a.
[tex]$\begin{align} (\cos x+\sin x)(\cos x-\sin x)&=1-2\sin^2 x \\ \cos^2x-\sin x\cos x+\sin x\cos x-\sin^2 x&=1-2\sin^2x \\ \cos^2x-\sin^2x&=1-2\sin^2x \\ (1-\sin^2x)-\sin^2x&=1-2\sin^2 x \\ 1-2\sin^2 x&=1-2\sin^2 x \\ Q.E.D\end{align}[/tex]

Bagian b.
[tex]$\begin{align} (\cos x+\sin x)^2&=1+2\cos x\sin x \\ \cos^2x+2\cos x\sin x+\sin^2 x&=1+2\cos x\sin x \\ (\sin^2x+\cos^2x)+2\cos x\sin x&=1+2\cos x\sin x \\ 1+2\cos x\sin x&=1+2\cos x\sin x \\ Q.E.D\end{align}[/tex]

7. turunan pertama dari fungsi y=cosx.sinx/cosx-sinxA. -1/(cosx+sinx)^2B. -2/(cosx+sinx)^2C -3/(cosx+sinx)^2D -1/(cosx-sinx)^2 tolong bantuannya


cepat ya
y ' = u'v - v'u / v²
v= cosx - sinx

di pilihan ganda hanya D saja. yang cosx - sinx
maka. jawabanya D


8. Buktikan 1-sinX/cosX + cosX/1-sinX = 2 secX


(1-sinX/cosX) + (cosX/1-sinX) = 2secX
(1-2sinX+sin^2X + cos^2X)/(cosX(1-sinX)) = 2secX
(1-2sinX+1)/(cosX(1-sinX)) = 2secX
(2-2sinX)/(cosX(1-sinX)) = 2secX
2(1-sinX)/(cosX(1-sinX)) = 2secX (coret (1-sinX))
2/cosX = 2secX
2secX = 2secX

"sin^2X + cos^2X = 1"Bab Trigonometri
Matematika SMA Kelas X

((1 - sin x) / cos x) + (cos x / (1 - sin x))
= ((1 - sin x)² + cos² x) / (cos x . (1 - sin x))
= (1 - 2 sin x + sin² x + cos² x) / ((1 - sin x) . (cos x))
= (1 - 2 sin x + 1) / ((1 - sin x) . cos x))
= (2 - 2 sin x) / ((1 - sin x) . (cos x))
= (2 (1 - sin x) / ((1 - sin x) . cos x))
= 2 / cos x
= 2 . sec x

terbukti


9. (sinx+cosx)2-(sinx-cosx)2=4sinx cos x



[tex]( {sinx + cosx)}^{2} - ( {sinx - cosx)}^{2} \\ =( {sin}^{2} x + 2sinxcosx + {cos}^{2} x) - ( {sin}^{2} x - 2sinxcosx + {cos}^{2} x) \\ = ( {sin}^{2} x + {cos}^{2} x + 2sinxcosx) - ( {sin}^{2} x + {cos}^{2} x - 2sinxcosx) \\ = (1 + 2sinxcosx) - (1 - 2sinxcosx) \\ = 1 + 2sinxcosx - 1 + 2sinxcosx \\ = 4sinxcosx \: \: \: terbukti[/tex]

10. buktikan bahwa: 1. sinx/1+cosx = 1-cosx/sinx 2. sinx/1-cosx= 1-cosx/sinx


1.
sinx / 1+cosx = 1-cosx / sinx
sinx (1 - cosx) / (1 + cosx)(1 - cosx) = 1-cosx/sinx
sinx (1 - cosx) / 1 - cos²x = 1-cosx / sinx
sinx (1 - cosx) / sin²x = 1-cosx / sinx
1-cosx / sinx = 1-cosx / sinx

2.
sinx / 1 - cosx = 1 + cosx / sinx
(1 - cosx)(1 + cosx) = sinx . sinx
1 - cos²x = sin²x
sin²x = sin²x

11. lim x mendekati π/2 sin(cosx)/cosx


Bab : Limit Fungsi Trigonometri

[tex] \lim_{x \to \frac{\pi}{2}} \frac{sin (cos x)}{cos x} [/tex]

Misal cos x = a

x → π/2

a → cos π/2

a → 0

[tex] \lim_{a \to 0} \frac{sin a}{a} [/tex]

Kita tahu bahwa bentuk tersebut limitnya adalah 1. Maka

[tex] \lim_{x \to \frac{\pi}{2}} \frac{sin (cos x)}{cos x} = 1 [/tex]


12. (cosx+Sony)^2/(cosx-sinx)


Penjelasan dengan langkah-langkah:

[tex] cos {}^{2} (x) + \sin {}^{2} (x) \div \cos(x) - \sin(x) = \cos(x) - \sin(x) [/tex]

maaf klw salah


13. Buktikan identitas trigojometri cosx^4-sinx^4=cosx^2-sinx^2


[tex]
\cos^4 x - \sin^4 x\\
=(\cos^2 x + \sin^2 x)( \cos^2 x - \sin^2 x)\\
= 1 ( \cos^2 x - \sin^2 x)\\
= \cos^2 x + \sin^2 x\\
\\
Q.E.D

[/tex]Ingat sifat
∴ a² - b² = (a-b)(a+b)
∴ sin² x + cos² x = 1

cos⁴ x - sin⁴ x
= (cos² x)² - (sin² x)²
= (cos² x - sin² x)(cos² x + sin² x)
= (cos² x - sin² x) * 1
= cos² x - sin² x
*) terbukti

14. tentukan integral tak tentu darin ((sinx/cosx) + (1/cosx))^2 tooollooong yah


∫ ( sin x/cos x  + 1/cos x)² dx
= ∫ (tan x + sec x)² dx
= ∫ tan² x + 2 sec x tan x + sec² x
= ∫ sec² x -1` + sec² x + 2 sec x tan x 
= ∫ 2 sec² x - 1 + 2 sec x tan x 
= ∫ 2 sec² x - ∫1 + ∫ 2sec x tan x 
= 2 tan x - x + 2 sec x + c

15. turunan pertama dari f(x) = (3x^2 - 5) cosx adalah a. 3x sinx + (3x^2 - 5) cosx b. 3x cosx + (3x^2 - 5) sinx c. -6x sinx - (3x^2 - 5) cosx d. 6x cosx + (3x^2 - 5) sinx e. 6x cosx - (3x^2 - 5) sinx


f(x) = u.v
f'(x) = u'v + v'u

misal u = 3x²-5 --> u'= 6x
v = cos x --> v' = - sin x

f'(x) = 6x (cos x) + (- sin x) (3x²-5)
= 6x cos x - (3x²-5) sin x

16. Sudut berapakah cosx=2


sin dikali 42d kali

17. intergal dari a. sinx^2b. -cosx^2c. cosx^-2pakek cara


a. sin²x = 1 - cos²x   ----> cos 2x = 2 cos²x -1
                                     2 cos²x = 1 + cos 2x
                                        cos²x = 1/2 + 1/2 cos 2x
 
  sin²x = 1 - (1/2 + 1/2 cos 2x)
          = 1/2 - 1/2 cos 2x
          
   
 [tex] \int\limits^._. { \frac{1}{2}- \frac{1}{2}cos2x } \, dx = \frac{1}{2}x- \frac{1}{4}sin2x+c [/tex]

b. - cos²x = - (1/2+1/2 cos 2x)
            
     [tex]- \int\limits^._. {\frac{1}{2}- \frac{1}{2}cos2x} \, dx =- \frac{1}{2}x+ \frac{1}{4}sin2x+c [/tex]

c. [tex]cos^{-2}x= \frac{1}{cos^{2}x} = \frac{1}{ \frac{1}{2} + \frac{1}{2}cos 2x} = \frac{2}{1+cos2x} [/tex]
 
    soal C perlu di konfirmasi 
            

18. jika cosx = 2 sinx, maka nilai sinx cosx adalah


[tex]\displaystyle \cos x=2\sin x\\\frac{\cos x}{\sin x}=2\\\cot x=2\\\frac{b}{a}=2\\\\b=2\\a=1\\c^2=a^2+b^2=1^2+2^2=1+4=5\\\\\sin x\cos x=\frac{a}{c}\times\frac{b}{c}\\\sin x\cos x=\frac{ab}{c^2}\\\sin x\cos x=\frac{1\times2}{5}\\\boxed{\boxed{\sin x\cos x=\frac{2}{5}}}[/tex]

19. jika y=x^2 cosx-2x sinx-2 cosx maka dy/dx adalah


Turunan Trigonometri

y = x² cos 2x - 2x sin x - 2 cos x

p = x² cos 2x --> p' = 2x cos  2x + x² (-2 sin 2x) = 2x cos x - 2x² sin 2x
q = 2x sin x --> q' = 2 sin x + 2x cos x
r = 2 cos x --> r' = - 2 sin x

y = p - q - r
dy/dx = y' = p' - q' - r'
y'=  2x cos x - 2x²  sin 2x - 2 sin x - 2x cos x + 2 sin x
y' =  - 2x² sin 2x


20. Cos (X-45°) = ..... A. 1/2√2(cosx + sinx) B. 1/2√2(cosx - sinx) C. 1/2√2(sinx - cosx) D. 1/2√3(cosx + sinx) E. 1/2√3(cosx - sinx)


Bab Trigonometri
Matematika SMA Kelas X

cos (x - 45°) = cos x . cos 45° + sin x . sin 45°
                     = cos x . 1/2 √2 + sin x . 1/2 √2
                     = 1/2 . √2 (cos x + sin x)

jawabannya A
Kategori matematika