# Proving

• Oct 21st 2009, 10:22 PM
pathurr19
Proving
I like proving but this one, I got stuck because of time pressure. It was the last question and I was hurrying since I only got 5 mins left. I didn't get to finish my solution so I may have gotten a few points off it.

here it goes:

1/cos^2B = [cosB sin2B - sinB cos2B]/cosB sinB

so what I did first was:

1/cos^2B = cos(B+2B)/cosB sinB

then I got stuck so I repeated it then came out with:

1/cos^2B = sinB - cosB/cosB sinB by subtracting the functions (idk if that's allowed) then I tried to break that down. I am still stuck.

anyone who can help me on how the sol'n goes??
• Oct 21st 2009, 10:46 PM
Hello pathurr19
Quote:

Originally Posted by pathurr19
I like proving but this one, I got stuck because of time pressure. It was the last question and I was hurrying since I only got 5 mins left. I didn't get to finish my solution so I may have gotten a few points off it.

here it goes:

1/cos^2B = [cosB sin2B - sinB cos2B]/cosB sinB

so what I did first was:

1/cos^2B = cos(B+2B)/cosB sinB

then I got stuck so I repeated it then came out with:

1/cos^2B = sinB - cosB/cosB sinB by subtracting the functions (idk if that's allowed) then I tried to break that down. I am still stuck.

anyone who can help me on how the sol'n goes??

I get a slightly different result.

$\frac{\cos B\sin2B-\sin B\cos2B}{\cos B \sin B}=\frac{\sin(2B-B)}{\cos B \sin B}$

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

$=\frac{1}{\cos B }$

• Oct 22nd 2009, 12:50 AM
reiward
Just like Grandad, I got different results.

$\frac{1}{cos^{2B}} = \frac{cosB sin2B - sinB cos2B}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{cosB (2sinB cosB) - sinB (2cos^{2}B - 1)}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{(2sinB cos^{2}B) - sinB (2cos^{2}B - 1)}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{(2sinB cos^{2}B) - (2sinBcos^{2}B - sin B)}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{sin B}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{sin B}{cosB sinB}$

$\frac{1}{cos^{2B}} = \frac{1}{cosB}$