SB19 Random Scenarios And OneShot (COMPLETED)

(125)SB19 as your math tutor

SB19 as your math tutor

📍Typographical and Grammatical Errors Ahead
📍Dedicated to all

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["JOSH"]

WHEN sin²x + cos²x = 1, tan²x + 1 = sec²x, 1 + cot²x = csc²x, (x, f(x)) sin 3xsin x x =sin x(2 cos2 x − 1)sin x cos x+cos x(2 sin x cos x)sin x cos x=2 cos2 x − 1cos x+ 2 cos x= 2 cos x −1cos x+ 2 cos x = 4 cos x − sec x dx/dy 2x³ + 4x² - 2x + 1, 1 + z²/dy/dx1 - z² 2dz ln(sinx) g(x) = 2x² sqrt(csc²x) y f(x) + g(x)√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)= mx + b f(x) = x² sin dy/dx =√1 − cos2 x =s1−2 32=√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos 2x cos x − sin 2x sin x= (2 cos x= 2 cos3 x − cos x − 2 cos x(1 − cos2 x)= 2 dy/dx x − sin 3x − sin xcos 3x + cos x=2 cos3x + x2sin3x − x22 cos3x + x2cos3x − x2=2 cos 2x sin x2 cos 2x cos x=sin xcos x= tan x
(f + g)(x) = f(x) + g(x)√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)• (f /sqrt(csc²x) y = mx + b f(x) = x² sin x =√1 − cos2 x =s1−2 32=√ = f(x)g(x), g(x) 6= 0• (f ◦ g)(x) = f(g(x))• : (f ◦ g)(x) to (g ◦ f)(x)sin A =BCAB =opposite hypotenuse cos A =ACAB =adjacent hypotenuse tan(A) = BCAC =opposite adjacent2 cos x −1cos x+ 2 cos x = 4 cos x − sec x dx/dy 2x³ + 4x² - dy/dx + 1, 1 + z²/1 - z² 2dz ln(sinx) g(x) = 2x² sqrt(csc²x) y = mx + b f(x) = x² sin x =√1 − cos2 x =s1−2csc A =1sin A=ABBC =hypotenuse oppositesec A =1cos A=ABAC
hypotenuse adjacentcot A =1tan A=ACBC = adjacent opposite1. sin(−x) = − sin x2. cos(−x) = cos x3. sin( π2 − x) = cos x4. cos( π2 − x) = sin x5. sin( π2 + x) = cos x6. cos( π2 + x) = − sin x7. sin(π − x) = sin x8. cos(π − x) = − cos x9. sin(π + x) = − sin x10. cos(π − x) = − cos x1. loga 1 = 02. loga a = 13. loga mn = loga m + loga n4. loga(f)
(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)• (f /sqrt(csc²x) y
cot²x = csc²x, (x, f(x)) sin 3xsin x x =sin x(2 cos2 x − 1)sin x cos x+cos x(2 sin x cos x)sin x cos x=2 cos2 x − 1cos x+ 2 cos x= 2 cos x −1cos x+ 2 cos x = 4 cos x − sec x dx/dy 2x³ + 4x² - 2x + 1, 1 + z²/dy/dx1 - z² 2dz ln(sinx) g(x) = 2x² sqrt(csc²x) y f(x) + g(x)√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)= mx + b f(x) = x² sin dy/dx =√1 − cos2 x =s1−2 32=√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos 2x cos x − sin 2x sin x= (2 cos x= 2 cos3 x − cos x − 2 cos x(1 − cos2 x)= 2 dy/dx x − sin 3x − sin xcos 3x + cos x=2 cos3x + x2sin3x − x22 cos3x + x2cos3x − x2=2 cos 2x sin x2 cos 2x cos x=sin xcos x= tan x
"cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)• (f /sqrt(csc²x) y = mx ."
(f + g)(x) = f(x) + g(x)√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)• (f /sqrt(csc²x) y = mx + b f(x) = x² sin x =√1 − cos2 x =s1−2 32=√ = f(x)g(x), g(x) 6= 0• (f ◦ g)(x) = f(g(x))• : (f ◦ g)(x) to (g ◦ f)(x)sin A =BCAB =opposite hypotenuse cos A =ACAB =adjacent hypotenuse tan(A) = BCAC =opposite adjacent2 cos x −1cos x+ 2 cos x = 4 cos x − sec x dx/dy 2x³ + 4x² - dy/dx + 1, 1 + z²/1 - z² 2dz ln(sinx) g(x) = 2x² sqrt(csc²x) y = mx + b f(x) = x² sin x =√1 − cos2 x =s1−2csc A =1sin A=ABBC =hypotenuse oppositesec A =1cos A=ABAC
"π − x) = − cos x1. loga 1 = 02. loga a =."
hypotenuse adjacentcot A =1tan A=ACBC = adjacent opposite1. sin(−x) = − sin x2. cos(−x) = cos x3. sin( π2 − x) = cos x4. cos( π2 − x) = sin x5. sin( π2 + x) = cos x6. cos( π2 + x) = − sin x7. sin(π − x) = sin x8. cos(π − x) = − cos x9. sin(π + x) = − sin x10. cos(π − x) = − cos x1. loga 1 = 02. loga a = 13. loga mn = loga m + loga n4. loga(f)
"cot²x = csc²x, (x, f(x)) sin 3xsin," x x =sin x(2 cos2 x − 1)sin x cos x+cos x(2 sin x cos x)sin x cos x=2 cos2 x − 1cos x+ 2 cos x= 2 cos x −1cos x+ 2 cos x = 4 cos x − sec x dx/dy 2x³ + 4x² - 2x + 1, 1 + z²/dy/dx1 - z² 2dz ln(sinx) g(x) = 2x² sqrt(csc²x) y f(x) + g(x)√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos• (f − g)(x) = f(x) − g(x)• (fg)(x) = f(x)g(x)= mx + b f(x) = x² sin dy/dx =√1 − cos2 x =s1−2 32=√53sin 3xsin xcos x= 4 cos x − sec x. cos 3x = cos(2x + x)= cos 2x cos x − sin 2x sin x= (2 cos x= 2 cos3 x − cos x − 2 cos x(1 − cos2 x)= 2 dy/dx x − sin 3x − sin xcos 3x + cos x=2 cos3x + x2sin3x − x22 cos3x + x2cos3x − x2=2 cos 2x sin x2 cos 2x cos x=sin xcos x= tan x

And they live mathematically ever after. ☺️🤧
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A/N: HAHAHAHAHA. HAPPY APRIL FOOLS DAAAY! SYEMPRE BABAWI AKO HAHAHA. Wait nyo SB19 scenario ko. Ipopost kona after kong mapost 'to. Love you all specially my kenstellationsss! 😽✨

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