# Given that:

• Sep 26th 2009, 08:58 AM
Sanjana Das
Given that:
$\displaystyle (1+cosA)(1+cosB)(1+cosY)=(1-cosA)(1-cosB)(1-cosY)$

Show tht one of the values of each member of this equality is sinA sinB sinY.

plss explain with all the steps as i m in 10th std.....
• Sep 26th 2009, 09:23 AM
Hello Sanjana Das
Quote:

Originally Posted by Sanjana Das
$\displaystyle (1+cosA)(1+cosB)(1+cosY)=(1-cosA)(1-cosB)(1-cosY)$

Show tht one of the values of each member of this equality is sinA sinB sinY.

plss explain with all the steps as i m in 10th std.....

Multiply both sides by $\displaystyle (1+\cos A)(1+\cos B)(1+\cos Y)$, noting that, on the RHS, $\displaystyle (1-\cos A)(1+\cos A) = (1-\cos^2A), ...$ etc:

$\displaystyle (1+\cos A)^2(1+\cos B)^2(1+\cos Y)^2=(1-\cos^2A)(1-\cos^2B)(1-\cos^2Y)$

Can you complete it now? (Hint: $\displaystyle 1-\cos^2A=\sin^2A$)

• Sep 26th 2009, 09:24 AM
Sanjana Das
Quote:

Hello Sanjana DasMultiply both sides by $\displaystyle (1+\cos A)(1+\cos B)(1+\cos Y)$, noting that, on the RHS, $\displaystyle (1-\cos A)(1+\cos A) = (1-\cos^2A), ...$ etc:
$\displaystyle (1+\cos A)^2(1+\cos B)^2(1+\cos Y)^2=(1-\cos^2A)(1-\cos^2B)(1-\cos^2Y)$
Can you complete it now? (Hint: $\displaystyle 1-\cos^2A=\sin^2A$)