(Trigonometry) Suppose sinA= 12/13 with 90 degrees

Question

(Trigonometry) Suppose sinA= 12/13 with 90 degrees < or equal to A < or equal to 180 degrees. Suppose also that sinB= -7/25 with -90 degrees < or equal to B < or equal to 0 degrees. Find cos(A + B). (Check picture incase I made typos) My textbook (online) doesn't show me the full steps of how to use the Pythagorean Theorem with sum and difference identites, so please walk me through ):

anonymous · Answer

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ajprincess · Answer

$$cosx=\sqrt{1-sin^2x}$$
$$cos(A+B)=cosAcosB-sinAsinB$$
can u do nw?

anonymous · Answer

That helps (: But what would I substitue A and B with? Like 30 degrees and 45 degrees since they are known on the Unit Circle?

ajprincess · Answer

No need substitute A and B with angles bcos u r given the value of sinA and sinB directly. u can use it to solve.
$$cosA=\sqrt{1-sin^2A}$$
Nw can u do?

anonymous · Answer

Ooooh, right Lol. Sorry. I think so? I'll try and if not I'll come back here

ajprincess · Answer

Oh k.

anonymous · Answer

So when I fill in for A in that equation would it look like this?
$$\cos 12/13 = \sqrt{1-\sin ^{2}(12/13)}$$

anonymous · Answer

I'm not really following, I'm sorry ):

anonymous · Answer

I think what's really throwing me off is the whole $$90 \le A \le180$$ thing

ajprincess · Answer

no. 
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that's jst an in4 sayng A lies between 90 nd 180.
Nthng to worry abt t.
can u do nw?
Sorry gotta go nw.