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Please help. Thank you in advance.
Given tan 𝛼 =7/24, 𝛼 in the third quadrant, sin 𝛽 =2/√13, 𝛽 in the second quadrant, find:
a) the quadrant containing 𝛼 + 𝛽
b) the quadrant containing 𝛼 - 𝛽
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Notice that 7^2 + 24^ = 625 = 25^2.
So, you may think that
tan(a) = in the third quadrant is = .
In other words, you may think that sin(a) = , cos(a) = .
The signs at sin(a) and cos(a) do agree that "a" is in the third quadrant.
Next, you are given that sin(b) = in the second quadrant; so, you can calculate
cos(b) = = = = = .
The sign at cos(b) does agree that "b" is in the second quadrant.
Now, as you know sin(a) , cos(a) = , sin(b) = , cos(b) = , you can calculate
sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b) = + = - =
= = ,
cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) = - = - =
= = .
Thus, sin(a+b) is a negative real number; cos(a+b) is a positive real number.
It means that angle a+b is in fourth quadrant.
ANSWER. Angle a+b is in fourth quadrant.
Part (a) is solved completely.
For part (b), calculate sin(a-b) and cos(a-b) similarly; then make a conclusion about angle a-b.
You just have a TEMPLATE for it.