trigonometry problems

**wintersoltice** · August 16th 2008, 03:49 AM

1.[SOLVED]
Given that $\frac{cos(A-B)}{cos(A+B)}$ = $\frac{7}{5}$ , 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= $45^\circ$ , 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 .

**kalagota** · August 16th 2008, 04:12 AM

Originally Posted by wintersoltice

1.
Given that $\frac{cos(A-B)}{cos(A+B)}$ = $\frac{7}{5}$ , 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= $45^\circ$ , calculate the value of tan A + tan B.

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

$\cos (A+B) = \cos (A)\cos (B) - \sin (A) \sin (B)<br />$

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

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

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

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 .

$A+B+C = \pi$ so, $A = \pi - (B+C)$ ...

**mr fantastic** · August 16th 2008, 04:17 AM

Originally Posted by wintersoltice

1.
Given that $\frac{cos(A-B)}{cos(A+B)}$ = $\frac{7}{5}$ , 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= $45^\circ$ , calculate the value of tan A + tan B.

Mr F says: ${\color{red}\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}}$ . 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: ${\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C }$ ${\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }$ ......

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 ......

..

**wintersoltice** · August 16th 2008, 06:07 AM

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......

**vishalgarg** · August 16th 2008, 06:12 AM

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....

**vishalgarg** · August 16th 2008, 06:17 AM

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....

**wintersoltice** · August 16th 2008, 06:24 AM

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 ....

**mr fantastic** · August 16th 2008, 03:13 PM

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: ${\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C }$ ${\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }$

[snip]

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

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

**kalagota** · August 16th 2008, 06:27 PM

Originally Posted by kalagota

$A+B+C = \pi$ so, $A = \pi - (B+C)$ ...

an easier approach is..

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

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

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

**mr fantastic** · August 16th 2008, 07:41 PM

Originally Posted by kalagota

an easier approach is..

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

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

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

You say tomayto, I say tomarto

..... The technique is obvious enough that the OP can decide which way to go.

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