Question

THS IS FUNCTIONS (CHAPTER ;RELATION BETWEEN SIDES AND ANGle or using sine formula show that each of the following holdand the law of sine rule

Q1 .tan A-B/2/TAN A+B/2=a-b/a+b
Q2 .b cos B+c cos C=a cos (B-C)
Q3 a sin B-C/2\=(b-c) cos A/2
Q4 b+c/b-c=tan B+C/2 * COT B-C/2
Q5 a cosA+b cos C+c cos C=2a sinB sinC
Q5. IN ANY TRIANGLE IF a/ cosA=b/cosB, PROVE THAT THE TRIANGLE IS ISOSCELES

Answer 1

Answer on Question #44504 – Math - Trigonometry

Problem.

THIS IS FUNCTIONS (CHAPTER ; RELATION BETWEEN SIDES AND ANGLE) or using sine formula show that each of the following hold and the law of sine rule

Q1. tan A-B/2/TAN A+B/2=a-b/a+b

Q2. b cos B+c cos C=a cos (B-C)

Q3. a sin B-C/2\=(b-c) cos A/2

Q4. b+c/b-c=tan B+C/2 * COT B-C/2

Q5. a cos A+b cos C+c cos C=2a sin B sin C

Q5. IN ANY TRIANGLE IF a/ cos A = b/cos B, PROVE THAT THE TRIANGLE IS ISOSCELES

Remark.

The statement isn't correctly formatted. I suppose that the correct statement is

"Q1. $\frac{\tan\frac{A - B}{2}}{\tan\frac{A + B}{2}} = \frac{a - b}{a + b}$

Q2. $b \cos B + c \cos C = a \cos (B - C)$

Q3. $a \sin \frac{B - C}{2} = (b - c) \cos \frac{A}{2}$

Q4. $\frac{b + c}{b - c} c = \tan \frac{B + C}{2} \cdot \cot \frac{B - C}{2}$

Q5. $a \cos A + b \cos B + c \cos C = 2a \sin B \sin C$ (I suppose that should be $b \cos B$ instead of $b \cos C$)

Q6. In any triangle if $\frac{a}{\cos A} = \frac{b}{\cos B}$, then the triangle is isosceles"

Solution.

Suppose that $a, b, c$ are the sides of triangle, $A, B, C$ are the degree measures of the angles and $R$ is the radius of circumcircle. By the law of sines

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R.$$

Therefore $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$.

Q1

$$\frac{a - b}{a + b} = \frac{2R \sin A - 2R \sin B}{2R \sin A + 2R \sin B} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{2 \sin \frac{A - B}{2} \cos \frac{A + B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \frac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}}$$

Q2

$$\begin{array}{l} b \cos B + c \cos C = 2R \sin B \cos B + 2R \sin C \cos C = R (\sin 2B + \sin 2C) \\ = 2R \sin (B + C) \cos (B - C) \\ = 2R \sin (\pi - A) \cos (B - C) = 2R \sin A \cos (B - C) = a \cos (B - C) \end{array}$$

Q3

$$\begin{array}{l} a \sin \frac{B - C}{2} = 2R \sin A \sin \frac{B - C}{2} = 2R \sin (\pi - B - C) \sin \frac{B - C}{2} = 4R \sin (B + C) \sin \frac{B - C}{2} \\ = 4R \sin \frac{B + C}{2} \cos \frac{B + C}{2} \sin \frac{B - C}{2} = 2R \cos \frac{\pi - B - C}{2} (\sin B - \sin C) \\ = \cos \frac{A}{2} (2R \sin B - 2R \sin C) = \cos \frac{A}{2} (b - c). \end{array}$$

Q4 From Q1 if $b \neq c$

$$\frac {b + c}{b - c} = \frac {\tan \frac {B + C}{2}}{\tan \frac {B - C}{2}} = \tan \frac {B + C}{2} \cdot \cot \frac {B - C}{2}.$$

Q5 From Q2

$$\begin{array}{l} a \cos A + b \cos B + c \cos C = a \cos A + a \cos (B - C) = a (\cos (\pi - B - C) + \cos (B - C)) \\ = a \left(\cos (B - C) - \cos (B + C)\right) = - 2 a \sin B \sin - C = 2 a \sin B \sin C. \\ \end{array}$$

Q6 $\frac{a}{\cos A} = \frac{b}{\cos B}$ if and only if $\frac{2R\sin A}{\cos A} = \frac{2R\sin B}{\cos B}$ or $\tan A = \tan B$ . $\tan A = \tan B$ if and only if $A = B$ , as $\tan x$ has period $\pi$ .

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