Math Help - exact value of cos75, sin15, cot-210

1. 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

2. 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 $\cos{(a-b)} = \cos{b}\cdot{\cos{a}} + \sin{b}\cdot{\sin{a}}$ works just fine. Let a = 45 degrees and b = 30 degrees.
Then $\cos{15} = \frac{\sqrt{2}}{2}\cdot{\frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2}\cdot{\frac{1}{2}}$.

3. 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 $\cos{(a+b)} = \cos{a}\cos{b} - \sin{a}\sin{b}$ for the computation of $\cos{75}$. And just so you're clear, $\cos{75} = \sin{15}$.

4. 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

5. hmm ok..im having trouble with csc(-225)

can people not just give tips but do one out please..visuals help me

6. $\left( { - 225^ \circ } \right) \cong \left( {135^ \circ } \right)$
If one goes 225degrees clockwise it is the same as going 135degrees counter-clockwise from the positive x-axis.

7. 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

8. 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.
$\csc ( - 225^ \circ ) = \csc (135^ \circ ) = \frac{1}{{\sin (135^ \circ )}}$