SOLUTION: cos(a) =-4/5 and a is in quadrant II and sin(b)=5/13 and b is in quadrant II. Find cos(a+b)

Question 1171626: cos(a) =-4/5 and a is in quadrant II and sin(b)=5/13 and b is in quadrant II. Find cos(a+b)

Found 3 solutions by Solver92311, ikleyn, MathTherapy:
Answer by Solver92311(821)   (Show Source): You can put this solution on YOUR website!

For an angle in QII:

For an angle in QII:

Then

You get to do your own arithmetic.

John

My calculator said it, I believe it, that settles it

From
I > Ø

Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.

For angle " a " in QII, sin(a) = = = = = (positive value). For angle " b " in QII, cos(b) = - = - = - = - = - (negative value). Now use cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b). Substitute the values into the formula and calculate.

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    - Evaluating trigonometric expressions
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Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
cos(a) =-4/5 and a is in quadrant II and sin(b)=5/13 and b is in quadrant II. Find cos(a+b)

cos (a + b) = cos (a) cos (b) - sin (a) sin (b)
From above, it's obvious that we need cos (b) and sin (a).
Given , and that ∡a is in the 2nd quadrant, we should realize that we're dealing with a 3-4-5 PYTHAGOREAN TRIPLE, which means that .
Likewise, given , and that ∡b is in the 2nd quadrant, we should realize that we're dealing with a 5-12-13 PYTHAGOREAN TRIPLE, which means that .
We now have:
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