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)

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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) About Me  (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(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.

For angle  " a "  in QII, 


    sin(a) = sqrt%281-cos%5E2%28a%29%29 = sqrt%281-%28-4%2F5%29%5E2%29 = sqrt%281-16%2F25%29 = sqrt%289%2F25%29 = 3%2F5  (positive value).



For angle  " b "  in QII, 


    cos(b) = - sqrt%281-sin%5E2%28b%29%29 = - sqrt%281-%285%2F13%29%5E2%29 = - sqrt%281-25%2F169%29 = - sqrt%28144%2F169%29 = - 12%2F13  (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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To see many other similar solved problems on calculating trig functions,  look into the lessons
    - Calculating trigonometric functions of angles
    - Advanced problems on calculating trigonometric functions of angles
    - Evaluating trigonometric expressions
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Trigonometry: Solved problems".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.



Answer by MathTherapy(10555) About Me  (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 matrix%281%2C3%2C+cos+%28a%29%2C+%22=%22%2C+%28-+4%29%2F5%29, 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 matrix%281%2C3%2C+sin+%28a%29%2C+%22=%22%2C+3%2F5%29.

Likewise, given matrix%281%2C3%2C+sin+%28b%29%2C+%22=%22%2C+5%2F13%29, 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 matrix%281%2C3%2C+cos+%28b%29%2C+%22=%22%2C+%28-+12%29%2F13%29.

We now have: