SOLUTION: 2 acute angles, A and B are such that cot A = 7 and tan(A - B) = -1. Without evaluating A and B, i) show that tan B = 4/3, ii) evaluate sin A and cos B, iii) evaluate sin^2(2A

Algebra ->  Trigonometry-basics -> SOLUTION: 2 acute angles, A and B are such that cot A = 7 and tan(A - B) = -1. Without evaluating A and B, i) show that tan B = 4/3, ii) evaluate sin A and cos B, iii) evaluate sin^2(2A      Log On


   

Question 1078891: 2 acute angles, A and B are such that cot A = 7 and tan(A - B) = -1.
Without evaluating A and B,
i) show that tan B = 4/3,
ii) evaluate sin A and cos B,
iii) evaluate sin^2(2A) + cos^2(2B).
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

Hello,
I have found that often when I do a hard problem like this for 
a student, the student often just turns in what I did and learns 
nothing.  So what I'm doing here is changing the numbers and 
doing a problem exactly like yours, that requires exactly the same 
steps from beginning to end.  So I'm doing this one instead of yours,
which only has the numbers changed.  Use it as a model to do
yours by, and you will learn something that way.  If you have
any trouble, or don't understand something, just tell me in the
thank-you note form below and I'll get back to you promptly by
email.  No charge ever, as I do this for fun.
2 acute angles, A and B, are such that cot A = 12 and tan(A-B) = -2.
Without evaluating A and B,
i) show that tan B = 5/2,
ii) evaluate sin A and cos B,
iii) evaluate sin^2(2A) + cos^2(2B).
cot%28A%29+=+1%2Ftan%28A%29=12 therefore tan%28A%29=1%2F12

tan%28A-B%29=%28tan%28A%29-tan%28B%29%29%2F%281%2Btan%28A%29tan%28B%29%29=-2

%28tan%28A%29-tan%28B%29%29%2F%281%2Btan%28A%29tan%28B%29%29=-2

tan%28A%29-tan%28B%29=-2%281%2Btan%28A%29tan%28B%29%29

tan%28A%29-tan%28B%29=-2-2tan%28A%29tan%28B%29%29

1%2F12-tan%28B%29=-2-2%281%2F12%29tan%28B%29

1%2F12-tan%28B%29=-2-%281%2F6%29tan%28B%29

Multiply thru by 12

1-12tan%28B%29=-24-2tan%28B%29

-10tan%28B%29=-25

tan%28B%29=%28-25%29%2F%28-10%29

tan%28B%29=5%2F2
ii) evaluate sin A and cos B,
cot%28A%29=12=12%2F1=adjacent%2F%28opposite%29

Draw a right triangle in QI that has angle A with adjacent=12
and opposite=1. Using the Pythagorean theorem, the hypotenuse 
is sqrt%2812%5E2%2B1%5E2%29=sqrt%28144%2B1%29=sqrt%28145%29. 

 

So sin%28A%29=opposite%2F%28hypotenuse%29=1%2Fsqrt%28145%29=sqrt%28145%29%2F145

-------------------

tan%28B%29=5%2F2=opposite%2F%28adjacent%29

Draw a right triangle in QI that has angle B with opposite=5
and adjacent=2. Using the Pythagorean theorem, the hypotenuse 
is sqrt%285%5E2%2B2%5E2%29=sqrt%2825%2B4%29=sqrt%2829%29. 

 

So cos%28B%29=adjacent%2F%28hypotenuse%29=2%2Fsqrt%2829%29=2sqrt%2829%29%2F29

-------------------------------------------------------
iii) evaluate sin^2(2A) + cos^2(2B).
We use the formulas and the two triangles above:

sin%282theta%29=2sin%28theta%29cos%28theta%29 and
cos%282theta%29=cos%5E2%28theta%29-sin%5E2%28theta%29

sin%282A%29=2sin%28A%29cos%28A%29=2%281%2Fsqrt%28145%29%29%2812%2Fsqrt%28145%29%29=24%2F145 and




Now do yours step-by-step in every detail exactly like this one.

Edwin