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Quan A.
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Trigonometry
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Question:

Prove that: $$[cos(x) - sin(x) ][ cos(2x) - sin(2x) ] = cos(x) - sin(3x)$$

Quan A.

Right side: $$cos(x) - sin(3x)$$ = $$cos(x) - sin( x + 2x)$$ = $$cos(x) - sin(x)cos(2x) - cos(x)sin(2x)$$ Left side: $$[cos(x) - sin(x) ][ cos(2x) - sin(2x)]$$ = $$cos(x) cos(2x) - cos(x) sin(2x) - sin(x) cos(2x) + sin(x) sin(2x)$$ = $$cos(x)(1 - 2 sin2) + sin(x) 2 sin(x) cos(x) - cos(x) sin(2x) - sin(x) cos(2x)$$ = $$cos(x) - 2 cos(x) sin2 + 2 cos(x) sin2 - cos(x) sin(2x) - sin(x) cos(2x)$$ = $$cos(x) - cos(x) sin(2x) - sin(x) cos(2x)$$ = left side

Geometry
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Question:

ABC is a right triangle. $$\widehat{BAC}$$ equals 90 degrees. AH is perpendicular to BC ( H is on BC). $$\widehat{ABC}$$ is equal to 60 degrees. Find the size of $$\widehat{HAC}$$ ?

Quan A.

The sum of all angles in triangle ABH is equal to 180°, so: $$\widehat{ABH}$$ + $$\widehat{BAH}$$ + 90° = 180° Substitute angle ABH by 60 and solve for angle BAH: $$\widehat{BAH}$$ = 180 - 90 - 60 = 30° However, $$\widehat{BAH}$$ + $$\widehat{HAC}$$ = $$\widehat{BAC}$$ = 90° $$\widehat{HAC}$$= $$\widehat{BAC}$$ - $$\widehat{BAH}$$ = 90° - 30° = 60°

Calculus
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Question:

Calculate indefinite integral: $$\int \sin ^{2}{t}\,dt\,$$

Quan A.

$$\int \sin ^{2}{t}\,dt\,$$ = $$\int {{(\frac {1}{2}}-{\frac {\cos 2t}{2}})dt}$$ = $$\int {{\frac {1}{2}}dt}-\int {{\frac {\cos 2t}{2}}dt}$$ = $${\frac {t}{2}}-{\frac {\sin 2t}{4}}$$ + C

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