Question

If \[\cos \theta= 3/5 and \cos \beta=(-5/13)\] where theta is acute and \[\pi <\beta <\frac{ 3\pi }{ 2 }\] find\[\cos(\theta+\beta)\]

Answer 1

Use cos(a+b) =cos(a) cos(b) - sin(a) sin(b)

Answer 2

okay, so I sub in the values in?

Answer 3

\(\theta\) is not delta.
Yes. Sub the values in.

If \(\cos \theta = \frac{3}{5} \implies \sin \theta = \frac{4}{5}\)

Also, \(\cos \beta\) should be \( = \frac{-5}{13}\) because it is in the 3rd quadrant (where cosine angles are negative)

Answer 4

oh oops, sorry

Answer 5

Also by delta I guess you mean theta... theta is acute means theta is between 0 and 90 which means sin(theta) and cos(theta) are both positive. Beta between 180 and 270 means both cos(beta) and sin(beta) are both negative you can use pythagoren identities to find both sin(theta) and sin(beta)... Also sorry for slowness not at home and typing this all on my phone.

Answer 6

So it would be cos(3/5)cos(-5/13)-sin(4/5)sin(12/13)??

Answer 7

As \(\beta\) is in the third quadrant, sin and cos both will always be negative.
\(\cos \beta = \frac{-5}{13}\) also, \(\sin^2 \beta + \cos^2 \beta = 1\)
\[\sin \beta = \pm \sqrt{1 - cos^2 \beta}\]
We reject the positive value of \(\sin \beta\) because sine is always negative in the third quadrant. [\(\beta\) is in the 3rd quadrant]

Answer 8

Sorry, I'm really new at this concept. Can you please explain it simpler?

Answer 9

Oh the CAST thingy?

Answer 10

Yes. http://mathonline.wikidot.com/cast-rule