1. ## trigonometry problems

1.[SOLVED]
Given that $\displaystyle \frac{cos(A-B)}{cos(A+B)}$ = $\displaystyle \frac{7}{5}$, prove that cos A cos B = 6 sin A sin B and deduce a relationship between tan A and tan B. Given further that A+B=$\displaystyle 45^\circ$, calculate the value of tan A + tan B.

2.
If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.
Hence, prove that sin A - sin(B-C)= 2 cos B sin C .

2. Originally Posted by wintersoltice
1.
Given that $\displaystyle \frac{cos(A-B)}{cos(A+B)}$ = $\displaystyle \frac{7}{5}$, prove that cos A cos B = 6 sin A sin B and deduce a relationship between tan A and tan B. Given further that A+B=$\displaystyle 45^\circ$, calculate the value of tan A + tan B.
$\displaystyle \cos (A-B) = \cos (A)\cos (B) + \sin (A) \sin (B)$

$\displaystyle \cos (A+B) = \cos (A)\cos (B) - \sin (A) \sin (B)$

so $\displaystyle 5 \cos (A-B) = 7\cos (A+B)$ ...

after that, $\displaystyle \frac{6\sin (A) \sin (B)}{\cos (A) \cos(B)} = 1$...

and $\displaystyle \tan (A + B) = \frac{\tan (A) + \tan (B)}{1 - \tan(A)\tan(B)}$

Originally Posted by wintersoltice
2.
If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.
Hence, prove that sin A - sin(B-C)= 2 cos B sin C .
$\displaystyle A+B+C = \pi$ so, $\displaystyle A = \pi - (B+C)$...

3. Originally Posted by wintersoltice
1.
Given that $\displaystyle \frac{cos(A-B)}{cos(A+B)}$ = $\displaystyle \frac{7}{5}$, prove that cos A cos B = 6 sin A sin B

Mr F says: Substitute from the appropriate double angle formula in the numerator and denominator, cross multiply to get rid of fractions, and simplify.

and deduce a relationship between tan A and tan B.

Mr F says: Divide both sides of cos A cos B = 6 sin A sin B by cos A cos B .....

Given further that A+B=$\displaystyle 45^\circ$, calculate the value of tan A + tan B.

Mr F says: $\displaystyle {\color{red}\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}}$. And from the previous result you know what tan A tan B is equal to.

2.
If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.

Mr F says: Here's a start: $\displaystyle {\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C }$ $\displaystyle {\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }$ ......

Hence, prove that sin A - sin(B-C)= 2 cos B sin C .

Mr F says: It's not difficult to use the compound angle formulae to show that sin (B + C) = sin (B - C) + 2 sin C cos B ......
..

4. i still don't get it for question 2

A + B + C = 180
A=180 - B - C

sin A -sin(B+C)=0
sin A = sin (B+C)

LHS
= sin A
=sin (180 - B- C)
=sin(-B-C)
= - sin (B+C)

but that doesn't equal to RHS......

5. its very simple apply componendo n dividendo.
cos(a-b) = cosa.cosb + sina.sinb
cos(a+b) = cosa.cosb - sina.sinb
so by componendo n dividendo u will get the ans....

6. its very simple apply componendo n dividendo.
cos(a-b) = cosa.cosb + sina.sinb
cos(a+b) = cosa.cosb - sina.sinb
so by componendo n dividendo u will get the ans....

by that relation you will get tanA.tanB=1/6.
if A+B = 45 YOU CAN GET value of tanA+TanB=tAN45(1-TANA.TANB)
WHICH IS EQUAL TO 5/6 IF YOU PUT VALUE OF TANA.TANB IN ABOVE EQUATION....

7. Originally Posted by vishalgarg
its very simple apply componendo n dividendo.
cos(a-b) = cosa.cosb + sina.sinb
cos(a+b) = cosa.cosb - sina.sinb
so by componendo n dividendo u will get the ans....

by that relation you will get tanA.tanB=1/6.
if A+B = 45 YOU CAN GET value of tanA+TanB=tAN45(1-TANA.TANB)
WHICH IS EQUAL TO 5/6 IF YOU PUT VALUE OF TANA.TANB IN ABOVE EQUATION....

thanks for your help but i have solved question 1....
question 2......still can't get it ....

8. Originally Posted by wintersoltice
thanks for your help but i have solved question 1....
question 2......still can't get it ....
Originally Posted by wintersoltice
[snip]
2.
If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.

Mr F says: Here's a start: $\displaystyle {\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C }$ $\displaystyle {\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }$

[snip]
$\displaystyle \sin A = \sin [180 - B] \cos C - \sin C \cos [180 - B]$

Now substitute $\displaystyle \sin [180 - B] = \sin B$ and $\displaystyle \cos [180 - B] = - \cos B$ ....

9. Originally Posted by kalagota
$\displaystyle A+B+C = \pi$ so, $\displaystyle A = \pi - (B+C)$...
an easier approach is..

$\displaystyle \sin A - \sin (B+C) = \sin (\pi - (B+C)) - \sin (B+C) = ...$

and by reduction formula, for any angle $\displaystyle x$, $\displaystyle \sin (\pi - x) = \sin\pi \cos x - \sin x \cos\pi = - (-1)\sin x = \sin x$

hence, $\displaystyle \sin (\pi - (B+C)) =\sin (B+C)$...

10. Originally Posted by kalagota
an easier approach is..

$\displaystyle \sin A - \sin (B+C) = \sin (\pi - (B+C)) - \sin (B+C) = ...$

and by reduction formula, for any angle $\displaystyle x$, $\displaystyle \sin (\pi - x) = \sin\pi \cos x - \sin x \cos\pi = - (-1)\sin x = \sin x$

hence, $\displaystyle \sin (\pi - (B+C)) =\sin (B+C)$...
You say tomayto, I say tomarto ..... The technique is obvious enough that the OP can decide which way to go.