Course Content
第四章:部分分式 Partial Fraction
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第六章:角的形成及单位 Angles and Measurements
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第十三章: 方程组 Simultaneous Equations
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第十五章:二元一次不等式及线性规划 linear inequality in two variables and linear programming
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高中一 | 高级数学
About Lesson
9.1 Basic Trigonometric Identities 基本三角关系式
1. Prove that 试证

(a) cos^4x - sin^4x = cos^2x - sin^2x

(b)

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(c) \frac{sec x-cosecx }{tan x - cot x} = \frac{tan x + cot x}{sec x + cosec x}
(d) \frac{{sin}^2{x}-{cos}^2{x}}{1+2sin{x}cos{x}}=\frac{tan{x}-1}{tan{x}+1}
(e) \frac{cos{(}-x)-1}{sin{(}3\pi+x)}=\frac{1}{cos{e}c(\pi-x)-tan{(}x-\frac{3\pi}{2})}
(f) \frac{2(cos{x}+1)}{sec{x}+tan{x}+1}=1-sin{x}+cos{x}
(g) \frac{tan{(}\pi+x)+cot{(}\frac{3\pi}{2}-x)}{1+tan{(}2\pi-x)cot{(}\frac{\pi}{2}+x)}=2sin{x}cos{x}
(2)
If u=\frac{1+sin{x}}{cos{x}}, prove that \frac{1}{u}=\frac{1-sin{x}}{cos{x}}

u=\frac{1+sin{x}}{cos{x}}, 证明 \frac{1}{u}=\frac{1-sin{x}}{cos{x}}

(3)
If tan^2A - 2tan^2B = 1, show that 2cos^2A - cos^2B = 0

tan^2A - 2tan^2B = 1, 证明 2cos^2A - cos^2B = 0

(4)
If sin A + cos A = \frac{1}{2} , find the values of
sin A + cos A = \frac{1}{2} , 求下列各式的值
(a) sin A cos A
(b) tan A + cot A
Answer 答案:
(a) -\frac{3}{8}

(b) -2\frac{2}{3}

(5)
If tan x + cot x = 4\frac{1}{2}, then find the value of tan2x + secx cscx + cot2x
tan x + cot x = 4\frac{1}{2} 则求 tan2x + secx cscx + cot2x 之值
Answer 答案:
22\frac{3}{4}
(6)
Given that sin x + cos x = 1.2, find sin x -  cos x, if 0 < x < 45^{\circ}
已知 sin x + cos x = 1.2, 求 sin x -  cos x, 当 0 < x < 45^{\circ}
Answer 答案:
-\sqrt{0.56}
(7)
Given that msec{\theta}=1+tan{\theta} , nsec{\theta}=1-tan{\theta}, prove that m^2 + n^2 = 2.

已知 msec{\theta}=1+tan{\theta} , nsec{\theta}=1-tan{\theta}, 证明 m^2 + n^2 = 2.

(8)
If \frac{a}{sin{A}}=\frac{b}{cos{A}} show that sin{A}cos{A}=\frac{ab}{a^2+b^2}

\frac{a}{sin{A}}=\frac{b}{cos{A}} 证明 sin{A}cos{A}=\frac{ab}{a^2+b^2}