Given:

SinA= -4/5

CosA= 3/5

CosA= radical8/3

CosB= -1/3

So the question is: cos(A-B) =cosA*cosB+sinA*sin(B)

(3/5)(-1/3)+(-4/5)(radical8/3)

-3/15+(-4radical8/15)=-3-4radical8/15

is this right!

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- Jul 21st 2013, 12:37 PMTrimHondurasDoes this looks ok!
Given:

SinA= -4/5

CosA= 3/5

CosA= radical8/3

CosB= -1/3

So the question is: cos(A-B) =cosA*cosB+sinA*sin(B)

(3/5)(-1/3)+(-4/5)(radical8/3)

-3/15+(-4radical8/15)=**-3-4radical8/15**

is this right! - Jul 21st 2013, 01:57 PMtopsquarkRe: Does this looks ok!
Aside from the SinB typo and needing parenthesis this is correct.

The correct answer is (-3 - 4radical8)/15.

Using LaTeX this is $\displaystyle \frac{-3 - 4 \sqrt{8}}{15}$. (See our LaTeX Help Forum!)

You can also simplify a bit. What is $\displaystyle \sqrt{8}$ in terms of $\displaystyle \sqrt{2}$?

-Dan