3 trig probs. evaluate cos, triangle,&addition formula

[letsgetaway]

I need help checking my methods of finding my answers. Thanks

1. Evaluate cos (7/6pi) + cos(-7/6pi). My answer:

-sqrt 3/2 + (-sqrt 3/2). Does this answer cancel out or can I add them?

2. ABC is a triangle with AB=2, BC=8, and AC=1. Find cos (A).

I have no idea how to find cos (A).

3. Simplify the expression sin (x+y) - sin (x-y).

sin (x+y) - sin (x+y) I distributed the negative sign.
sin x cos y + cos x sin y - ( sin x cos y + cos y sin x)
= 0

 

[tkhunny]

1) Why would summing two equal values "cancel out"?

1+1 = 2
-1 + (-1) = -2

2) Have you considered the "Law of Cosines"?

3) You did what with the negative sign? That would be the "Distributive Property of Multiplication over Addition". The sine function is NOT multiplication.

 

[pka]

Re: 3 trig probs. evaluate cos, triangle,&addition formu

1. Evaluate cos (7/6pi) + cos(-7/6pi).

It should be noted that cosine is an even function.
Therefore, \(\displaystyle \L
\cos ( - \theta ) = \cos (\theta )\quad \Rightarrow \quad \cos \left( {\frac{{7\pi }}{6}} \right) = \cos \left( {\frac{{ - 7\pi }}{6}} \right)\)

 

[letsgetaway]

Ok. I definitely gotta work these kinks out before Cal I. I'm so confused by some trig identities and triangles.