SOLUTION: Prove cot(45°-A) = {{{ (cotA + 1)/ (cotA - 1) }}} Hence show that cot15° = 2 + sqrt3

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: Prove cot(45°-A) = {{{ (cotA + 1)/ (cotA - 1) }}} Hence show that cot15° = 2 + sqrt3       Log On

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Question 217335: Prove cot(45°-A) = +%28cotA+%2B+1%29%2F+%28cotA+-+1%29+
Hence show that cot15° = 2 + sqrt3
Answer by jsmallt9(2063) About Me  (Show Source):
You can put this solution on YOUR website!
cot(45°-A) = +%28cotA+%2B+1%29%2F+%28cotA+-+1%29+

There may be more clever ways of doing this but one approach to a lot of Trig. problems is to start by rewriting the expression in terms of sin and cos. Since cot(a) = cos(a)/sin(a):
cot(45°-A) = +cos%2845+-A%29%2Fsin%2845-A%29+
Now we can use the angle difference identitites: cos%28x-y%29+=+cos%28x%29%2Acos%28y%29+%2B+sin%28x%29%2Asin%28y%29 and sin%28x-y%29+=+sin%28x%29%2Acos%28y%29+-+cos%28x%29sin%28y%29 to rewrite the right side:
cot(45°-A) = %28cos%2845%29cos%28A%29+%2B+sin%2845%29sin%28A%29%29%2F%28sin%2845%29cos%28A%29-cos%2845%29sin%28A%29%29
Since cos%2845%29+=+sin%2845%29+=+sqrt%282%29%2F2:
cot(45°-A) = %28%28sqrt%282%29%2F2%29cos%28A%29+%2B+%28sqrt%282%29%2F2%29sin%28A%29%29%2F%28%28sqrt%282%29%2F2%29cos%28A%29-%28sqrt%282%29%2F2%29sin%28A%29%29
We can reduce the fraction by factoring out sqrt%282%29%2F2:
cot(45°-A) = %28%28sqrt%282%29%2F2%29%28cos%28A%29+%2B+sin%28A%29%29%29%2F%28%28sqrt%282%29%2F2%29%28cos%28A%29-sin%28A%29%29%29
Now the %28sqrt%282%29%2F2%29's cancel leaving:
cot(45°-A) = %28cos%28A%29+%2B+sin%28A%29%29%2F%28cos%28A%29-sin%28A%29%29
Looking at what we have and at where we want to be we can see that we're very close. All we need is sin(A) as a denominator of each term. So we multiply the top and bottom by 1%2Fsin%28A%29:
cot(45°-A) = %28%281%2Fsin%28A%29%29%2F%281%2Fsin%28A%29%29%29%28%28cos%28A%29+%2B+sin%28A%29%29%2F%28cos%28A%29-sin%28A%29%29%29
Using the Distributive Property we get:
cot(45°-A) = %28cos%28A%29%2Fsin%28A%29+%2B+sin%28A%29%2Fsin%28A%29%29%2F%28cos%28A%29%2Fsin%28A%29+-+sin%28A%29%2Fsin%28A%29%29
which simplifies to
cot(45°-A) = +%28cot%28A%29+%2B+1%29%2F+%28cot%28A%29+-+1%29+

To find cot(15°): If we realize that 15 = 45-30 then we can use the formula above by setting A = 30:
cot(15°) = cot(45°-30) = +%28cot%2830%29+%2B+1%29%2F+%28cot%2830%29+-+1%29+
cot%2830%29+=+cos%2830%29%2Fsin%2830%29+=+%28sqrt%283%29%2F2%29%2F%281%2F2%29+=+sqrt%283%29
Substituting this we get:
cot(15°) = cot(45°-30) = +%28sqrt%283%29+%2B+1%29%2F+%28sqrt%283%29+-+1%29+
Next we rationalize the denominator. To do this we will use the conjugate of the denominator which is sqrt%283%29+%2B+1:
cot(15°) = cot(45°-30) = %28sqrt%283%29+%2B+1%29%2F%28sqrt%283%29+%2B+1%29%29%28sqrt%283%29+%2B+1%29%2F+%28sqrt%283%29+-+1%29+
Multiplying this out (using either FOIL or the patterns for (a+b)(a+b) and (a+b)(a-b)) we get:
cot(15°) = cot(45°-30) = %28%28sqrt%283%29%29%5E2+%2B+2sqrt%283%29+%2B+1%5E2%29%2F%28%28sqrt%283%29%29%5E2+-+1%5E2%29%29
cot(15°) = cot(45°-30) = %283+%2B+2sqrt%283%29+%2B+1%29%2F%283+-+1%29
See how the conjugate made the denominator rational?
cot(15°) = cot(45°-30) = %284+%2B+2sqrt%283%29%29%2F%282%29
Reduce the fraction by factoring out 2 and canceling:
cot(15°) = cot(45°-30) = %282%282+%2B+sqrt%283%29%29%29%2F%282%29
cot(15°) = cot(45°-30) = 2+%2B+sqrt%283%29