9.2 两角和与差的三角函数 Compound Angles - 独中课程 | 线上补习

高中一 | 高级数学

About Lesson

(1)
Given that $sin{A}=\frac{2}{3}$ , where $\frac{\pi}{2}<A<\pi$ ; $cos{B}=-\frac{3}{4}$ , where $\pi<B<\frac{3\pi}{2}$ , find the value of $cos({A} - {B})$ .
已知 $sin{A}=\frac{2}{3}$ , 且 $\frac{\pi}{2}<A<\pi$ ; , 且 $cos{B}=-\frac{3}{4}$ , 求 $cos({A} - {B})$ 的值.

Answer:

$\frac{3\sqrt5-2\sqrt7}{12}$

(2)
Prove that $tan{7}5^{\circ}=2+3$
证明 $tan{7}5^{\circ}=2+3$

(3)
Show that, if $A + B = \frac{\pi}{4}$ , $cot A - cot A tan B - tan B = 1$
试证，若 $A + B = \frac{\pi}{4}$ , $cot A - cot A tan B - tan B = 1$

(4)
If $tan x$ and $tan y$ are the roots of equation $x^2 - 7x - 5 = 0$ , find the value of $tan(x + y)$ .
若 $tan x$ 及 $tan y$ 为方程式 $x^2 - 7x - 5 = 0$ 的根, 求 $tan(x + y)$ 的值.

Answer:

$\frac{7}{6}$

(5)
If A + B = 90

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and A > 0

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, B > 0

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, prove that tan A tan B = 1.
Hence without using calculator or mathematical tables, find the value of tan 75

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– tan15

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.
若A + B = 90

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和 A > 0

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, B > 0

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, 证明 tan A tan B = 1.
据此不用计算机或查表，求 tan 75

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的值.

Answer:

$2\sqrt{3}$

Prove the following identities.
证明下列恒等式

(6)
$\frac{sin(A+B)}{cos{A}cos{B}}=tan{A}+tan{B}$

(7)
$\frac{sin(A-B)}{sin{A}sin{B}}=cot{B}-cot{A}$

(8)
$cos (A + B) cos (A - B) = cos^2A - sin^2B$

(9)
$sin (A + B) sin (A - B) = cos^2B - cos^2A$

(10)
$cos (45^{\circ} - A) - sin (45^{\circ} + A) = 0$

(11)
$cos (A - B) - sin (A + B) = (cosA - sinA)(cosB - sinB)$

(12)
$cos (A + B) cosB + sin (A + B) sinB = cosA$

(13)
$sin3A cosA - cos3A sinA = sin2A$

(14)
$cos (30^{\circ} + A) cos(30^{\circ} - A) - sin(30^{\circ} + A) sin(30^{\circ} - A) = \frac{1}{2}$

(15)
$tan{(}A+B)-tan{A}=\frac{sin{B}}{cos{A}cos{(}A+B)}$

(16)
Prove that in triangle $ABC, tan A + tan B + tan C = tan A tan B tan C$
试证在三角形ABC中, $tan A + tan B + tan C = tan A tan B tan C$