# Trignometry qus

• Aug 31st 2007, 07:42 PM
tom_asian
Trignometry qus
Find all values of x in the interval 0≤X2 pie

a) sin x = cos x

Could you please provide me working for it, thanks.
• Aug 31st 2007, 07:48 PM
tom_asian
Sorry, dont worry about working the qus out..

I just figured out how to do it, thanks
• Aug 31st 2007, 07:50 PM
Jhevon
Quote:

Originally Posted by tom_asian
Find all values of x in the interval 0≤X2 pie

a) sin x = cos x

Could you please provide me working for it, thanks.

well, you should pretty much know by heart that $\displaystyle \sin x = \cos x$ for $\displaystyle x = \frac {\pi}{4} + k \pi$

but if you wanted to work it out explicitly, here's what to do

$\displaystyle \sin x = \cos x$

$\displaystyle \Rightarrow \sin^2 x = \cos^2 x$

$\displaystyle \Rightarrow \cos^2 x - \sin^2 x = 0$

$\displaystyle \Rightarrow \cos 2x = 0$ ........you should also know for what values $\displaystyle \cos x = 0$

$\displaystyle \Rightarrow 2x = \frac {\pi}{2} + 2k \pi$

$\displaystyle \Rightarrow x = \frac {\pi}{4} + k \pi$

Now i leave the easy part for you. what are the values in the interval we were given?
• Aug 31st 2007, 07:51 PM
Jhevon
Quote:

Originally Posted by tom_asian
Sorry, dont worry about working the qus out..

I just figured out how to do it, thanks

couldn't you have said that sooner? :p
• Aug 31st 2007, 11:01 PM
topsquark
Quote:

Originally Posted by Jhevon
well, you should pretty much know by heart that $\displaystyle \sin x = \cos x$ for $\displaystyle x = \frac {\pi}{4} + k \pi$

but if you wanted to work it out explicitly, here's what to do

$\displaystyle \sin x = \cos x$

$\displaystyle \Rightarrow \sin^2 x = \cos^2 x$

$\displaystyle \Rightarrow \cos^2 x - \sin^2 x = 0$

$\displaystyle \Rightarrow \cos 2x = 0$ ........you should also know for what values $\displaystyle \cos x = 0$

$\displaystyle \Rightarrow 2x = \frac {\pi}{2} + 2k \pi$

$\displaystyle \Rightarrow x = \frac {\pi}{4} + k \pi$

Now i leave the easy part for you. what are the values in the interval we were given?

Ah! But $\displaystyle cos(2x) = 0$ for
$\displaystyle x = \frac{ \pi}{2} + 2k \pi \text{ and } \frac{3 \pi}{2} + 2k \pi$

So
$\displaystyle x = \frac{ \pi}{4} + k \pi \text{ and } \frac{3 \pi}{4} + k \pi$

It is just that the $\displaystyle 3 \pi /4 + k \pi$ solution is extraneous in this case, its origin coming from squaring both sides of the original equation.

-Dan
• Sep 1st 2007, 04:28 PM
ThePerfectHacker
And easier solution.
$\displaystyle \sin x = \cos x$
Divide by $\displaystyle \cos x$ to get,
$\displaystyle \tan x = 1$ now continue...