sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

pacman · #1 $Old$ October 9th, 2009, 01:58 AM

#. - If A, B, C are the angles of a triangle; show that
i) (tan A - cot B) = cos C sec A csc B.
ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

mathaddict · #2 $Old$ October 9th, 2009, 04:21 AM

Quote:

Originally Posted by pacman $View Post$

#. - If A, B, C are the angles of a triangle; show that

i) (tan A - cot B) = cos C sec A csc B.
ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

HI

For (1)

$\tan A -\cot B$

$=\frac{\sin A\sin B-\cos A\cos B}{\cos A\sin B}$

$=\frac{-\frac{1}{2}[\cos (A+B)-\cos (A-B)]-\frac{1}{2}[\cos (A+B)+\cos (A-B)]}{\cos A\sin B}$

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

$=\frac{\cos [180-(A+B)]}{\cos A\sin B}$

$=\frac{\cos C}{\cos A\sin B}$

= cos C sec A cosec B

Hence proved .

Grandad · #3 $Old$ October 9th, 2009, 10:45 PM

Hello pacman

Quote:

Originally Posted by pacman $View Post$

#. - If A, B, C are the angles of a triangle; show that
i) (tan A - cot B) = cos C sec A csc B.
ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

For part 2:

$\sin2A+\sin2B+\sin2C = 2\sin(A+B)\cos(A-B)+2\sin C\cos C$

$=2\sin C(\cos(A-B)+\cos C)$ , since $\sin(A+B) = \sin[180-(A+B)]=\sin C$

$=4\sin C\cos\tfrac12(A+C-B)\cos\tfrac12(A-B-C)$

$=4\sin C \cos\tfrac12([180-B]-B)\cos\tfrac12(A-[180-A])$

$=4\sin C \cos(90-B)\cos(A-90)$

$=4\sin C \cos(90-B)\cos(90-A)$

$=4\sin A \sin B\sin C$

Grandad

pacman · #4 $Old$ October 10th, 2009, 03:17 AM

Thanks grandad and mathaddict , i messed up the RHS, i did not think LHS is easier to work out but it is. Thanks

jeremyparker10 · #5 $Old$ October 11th, 2009, 02:05 PM

Left Hand Side (LHS) = 2 sin A. cos A + 2 sin (B+C). cos (B - C)

=> 2 sin A. cos A + 2 sin (180 - A).cos ( B - C )

=> 2 sin A. cos A + 2 sin A. cos ( B - C )

=> 2 sin A [ cos A + cos ( B- C ) ]

( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )

=> 2 sin A [ 2 sin B. sin C ]

=> 4. sin A. sin B. sin C ................... Proved