Thread: trigonometry proof

Show that in triangle ABC



I had tried to change cos A = cos (180 - (B+C) ) = - cos (b+C) = - cos B cos C + sin B sin C then stuck

Thanks

Originally Posted by songoku

Show that in triangle ABC



I had tried to change cos A = cos (180 - (B+C) ) = - cos (b+C) = - cos B cos C + sin B sin C then stuck

Thanks

Hi,

Using the cosine rule,

\cos C=\frac{a^2+b^2-c^2}{2ba}

\cos A = \frac{c^2+b^2-a^2}{2bc}

Find for cos B as well. Then substitute into the LHS of the equation and the remaining work is just algebra.

Originally Posted by mathaddict

Hi,

Using the cosine rule,

\cos C=\frac{a^2+b^2-c^2}{2ba}

\cos A = \frac{c^2+b^2-a^2}{2bc}

Find for cos B as well. Then substitute into the LHS of the equation and the remaining work is just algebra.

Wow, that's really great idea, never cross my mind. I've tried it and yes it worked, but my teacher didn't accept it because we are supposed to do it with trigonometry identity. Can you direct me how to use trig. identity to solve this?

Thanks

Originally Posted by songoku

Wow, that's really great idea, never cross my mind. I've tried it and yes it worked, but my teacher didn't accept it because we are supposed to do it with trigonometry identity. Can you direct me how to use trig. identity to solve this?

Thanks

True since this comes under trigonometry, my teacher wouldn't accept it either. WTell, i will try another way.

from the sine rule.

Lets begin from the LHS, divide both the denominator and numerator by a,















Hope this works.

Originally Posted by mathaddict

True since this comes under trigonometry, my teacher wouldn't accept it either. WTell, i will try another way.

from the sine rule.

Lets begin from the LHS, divide both the denominator and numerator by a,















Hope this works.

Unfortunately, it didn't. No sine rule or cosine rule. Phew....