If sin A= -4/5 with A in QIV, and sin B= 3/5 with B in QII, find cos(A+B)

Answer 1

Bella:

The best way to do this is to draw your two triangles in the proper quadrants. From there you can determine that in Quadrant II, the Cos(B) = -4/5 and in Quadra IV, the Cos(A) = +3/5 by the Pythagorean Thm or by noticing that we have a Pythagorean 3-4-5 triple. We already know that Sin(A) = -4/5 and Sin(B) = 3/5. Hopefully you used the Pythagorean Thm or the concept of a Pythagorean triple to determine the values not given to you. With all these 4 important trig functions, and by knowing the identity that

Cos(A+B) = CosACosB-SinASinB we get (3/5)(-4/5) - (-4/5)(3/5) or -12/25 - (-12/25) or -12/25 + 12/25 = 0

Answer 2

Hello Bella,

cos(A+B) = cosA*cosB - sinA*sinB (sum identity formula that should be in your class book/notes)

You're given sin A and sin B, so you need cos B and cos A.

You're told A is in QIV, so cos A will be POSITIVE. (#1)

You're told B is in QII, so cos B will be NEGATIVE. (#2)

Another fundamental trig formula you would need to know here: (sin x)^2 + (cos x)^2 = 1

Applying this for A: (sin A)^2 + (cos A)^2 = 1 so (cos A)^2 = 1 - (-4/5)^2 = 1 - 16/25 = 9/25.

That means: cos A = +/- square root of (9/25). Recall statement #1 from a few lines ago, cos A being positive means we choose the positive answer, so cos A = 3/5.

for B: (sin B)^2 + (cos B)^2 = 1 so (cos B)^2 = 1 - (3/5)^2 = 1-9/25 = 16/25

so cos B = +/- square root of (16/25) = +/- (4/5). Recall statement #2. B being in QII means a NEGATIVE cosine, so cos B = -4/5.

You now have all the 4 parameters you need: cos A = 3/5; cos B = -4/5; sin A = -4/5; sin B = 3/5

Plug them into the following formula: cos(A+B) = cosA*cosB - sinA*sinB

You will get the answer from there.

Cheers!