# cos45=sin45=1/2root2

• February 2nd 2009, 08:32 AM
mathmo
cos45=sin45=1/2root2
Sorry! Made a mistake in the first post... this is the correct question

By considering a right angled isosceles triangle, or otherwise, show that

1 + 2 sin 45
5 cos 45 - 2 sin 45

in the form a + b root 2 where the constants a and b are to be found...
• February 2nd 2009, 08:48 AM
RE :
Quote:

Originally Posted by mathmo
I dont understand this:

By considering a right angled isosceles triangle, or otherwise, show that

cos 45 = sin 45 = 1/2 root 2
5 cos 45 - 2 sin 45

in the form a + b root 2 where the constants a and b are to be found...

You can draw a right angle triangle with equal length of 1 and hypothenus $\sqrt{2}$ . From there , you can show that $cos{45}=sin{45}=\frac{1}{\sqrt{2}}$
$5cos{45}-2sin{45}=\frac{5}{\sqrt{2}}-\frac{2}{\sqrt{2}}=\frac{3}{\sqrt{2}}