Question 1058925: If you know that sin(a)=2/3 and cos(b)=-1/7, find the exact value of sin (a+b)
I used the equation sin(a+b)=sin(a)cos(b)+cos(a)+sin(b)
the way I understood how to do this, is that they were two different triangles(I do not know if they have to be two different triangles, or if it can be one triangle)
then I had sin(a+b)=(2/3)(-1/7)+((squareroot5)/3)(4*(squareroot3)/7)
simplified to (-2/21)+(4(*squareroot15)/21
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website! .
If you know that sin(a)=2/3 and cos(b)=-1/7, find the exact value of sin (a+b)
I used the equation sin(a+b)=sin(a)cos(b)+ + (cos(a)*sin(b)
the way I understood how to do this, is that they were two different triangles(I do not know if they have to be two different triangles, or if it can be one triangle)
then I had sin(a+b)=(2/3)(-1/7)+((squareroot5)/3)(4*(squareroot3)/7)
simplified to (-2/21)+(4(*squareroot15)/21
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Your solution (calculations) is (are) correct.
These are two angles. They are not the angles of one triangle. Not necessary.
The only thing you need to take an additional care is to correctly select the signs at the square roots.
For it, the problem should (must !) contain the info where (in which quadrants) the angles "a" and "b" are.
For many other similar solved problems see the lessons
- Calculating trigonometric functions of angles
- Advanced problems on calculating trigonometric functions of angles
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".
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This part of my post is the answer and the message to the tutor "MathTherapy".
It is a mistake to think that at the given conditions the angle "a" is in QII.
The angles "a" and "b" are INDEPENDENT angles and values, and there are no reasons to think that the angle "a" is in QII.
At given condition it may be in QI or QII, and there is no info to recognize it more precisely.
Answer by MathTherapy(10555)  (Show Source):
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