exact value of cos75, sin15, cot-210

• May 1st 2008, 04:03 PM
stones44
exact value of cos75, sin15, cot-210
ok so i need to find exact value of Cos75, Sin15 and Cot-210

for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

and i dont know how to do this...i keep getting the wrong answer

thanks
• May 1st 2008, 04:09 PM
icemanfan
Quote:

Originally Posted by stones44
ok so i need to find exact value of Cos75, Sin15 and Cot-210

for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

and i dont know how to do this...i keep getting the wrong answer

thanks

Your identity $\displaystyle \cos{(a-b)} = \cos{b}\cdot{\cos{a}} + \sin{b}\cdot{\sin{a}}$ works just fine. Let a = 45 degrees and b = 30 degrees.
Then $\displaystyle \cos{15} = \frac{\sqrt{2}}{2}\cdot{\frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2}\cdot{\frac{1}{2}}$.
• May 1st 2008, 04:13 PM
icemanfan
Quote:

Originally Posted by stones44
ok so i need to find exact value of Cos75, Sin15 and Cot-210

for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

and i dont know how to do this...i keep getting the wrong answer

thanks

You will probably want to use the identity $\displaystyle \cos{(a+b)} = \cos{a}\cos{b} - \sin{a}\sin{b}$ for the computation of $\displaystyle \cos{75}$. And just so you're clear, $\displaystyle \cos{75} = \sin{15}$.
• May 2nd 2008, 01:02 AM
Moo
Hello,

For the third one, you know that cot is an odd function.
Hence cot(-210)=-cot(210)

cot=cos/sin

Take 210=180+30
• May 2nd 2008, 02:36 PM
stones44
hmm ok..im having trouble with csc(-225)

can people not just give tips but do one out please..visuals help me
• May 2nd 2008, 02:45 PM
Plato
$\displaystyle \left( { - 225^ \circ } \right) \cong \left( {135^ \circ } \right)$
If one goes 225degrees clockwise it is the same as going 135degrees counter-clockwise from the positive x-axis.
• May 2nd 2008, 03:21 PM
stones44
well can you do one of them for me so i can get a feel for it? i get confused when there are multiple angles and stuff

mainly the csc one cause there arent identitys that ive learned that work with it
• May 2nd 2008, 03:55 PM
Plato
Quote:

Originally Posted by stones44
the csc one cause there arent identitys that ive learned that work with it

That is an impossible for anyone who is doing a serious study of trigonometric functions.
$\displaystyle \csc ( - 225^ \circ ) = \csc (135^ \circ ) = \frac{1}{{\sin (135^ \circ )}}$