Question

Given cos = -2/5 a in quadrant III and cos b = 1/4, b in quadrant I find

Sin(a+b)
Cos(a+b)
Tan(a+b)

Answer 1

I guess you mean
`\cos a=-\frac25`. Since
`a` is in quadrant III, we expect
`\sin a<0`. Then

`\sin a=-√(1-\cos^2a)=-\frac{√(21)}5`

Since
`b` is in quadrant I, we expect
`\sin b>0`, so that

`\sin b=√(1-\cos^2b)=\frac{√(15)}4`

Now,

`\sin(a+b)=\sin a\cos b+\cos a\sin b=-(2√(15)+√(21))/(20)`

`\cos(a+b)=\cos a\cos b-\sin a\sin b=(3√(35)-2)/(20)`

and

`\tan(a+b)=(\sin(a+b))/(\cos(a+b))=-(2√(15)+√(21))/(3√(35)-2)`