trig addition

[tsh44]

Hi I have this homewrok problem I would like to clear up. I got 7/SQRT(10) as the answer but I was told the answer should be -7SQRT(2)/10.

I am given these:

Sec A= 5/4
tan B= -1
0 <A<Pi/2<B<Pi

I had to find cos (A+B)

I first flipped Sec A to Cos A so I got 4/5 and then I figured the y value would be 3.

Then I worked on Tan B and found the y to be -1, the x to be 1, and the r value to be SQRT 2.

I also deduced that A is int he 1st Quad, and B is in the 2nd and started to plug in.

cosA Cos B- SinA SinB
[(-4/5) (-1/SQRT(2)] - (3/5) [(1/SQRT(2)]

So I got -4-3/ SQRT(10) as a final answer.
Can you tell me where I went wrong. Thank you.

 

[soroban]

Hello, tsh44!

Your work is correct (in fact, excellent!) . . . until the last step.

You had: \(\displaystyle \L\,\left(\frac{4}{5}\right)\left(-\frac{1}{\sqrt{2}}\right)\,-\,\left(\frac{3}{5}\right)\left(\frac{1}{\sqrt{2}}\right)\)

So I got \(\displaystyle \frac{-4\,-\,3}{\sqrt{10}}\) as a final answer.\(\displaystyle \;\;\) . . . no

We have: \(\displaystyle \L\:-\frac{4}{5\sqrt{2}} \,- \,\frac{3}{5\sqrt{2}} \;= \;-\frac{7}{5\sqrt{2}}\)


Rationalize: \(\displaystyle \L\:-\frac{7}{5\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;= \;-\frac{7\sqrt{2}}{10}\)

 

[tsh44]

Ohhhhhhh. Thank you. Pretty silly of me.