Expert: Sherry Wallin - 11/25/2009

Question
QUESTION: Question1 :If x and y are acute angles and sin2x =cosy, find x in terms of y.
Question2 :Determine the general solution for theta, if,
a) 7cos theta-2sin square theta +5 = 0
b) Hence solve the equation in the interval -720 degree less than or equal theta less than or equal 0 degree.

Thanks
Abram

ANSWER: Hi J. A~

For #1 I would do this:
take the arcsin of both sides:
arcsin(sin(2x)) = arcsin(cos(y))
2x = arcsin(cos(y))
x = arcsin(cos(y))/2

For #2 change sin^2 theta into 1-cos^2 theta using the identity cos^2 theta + sin^2 theta = 1
7 cos theta -2(1-cos^2 theta) + 5 = 0 -> 2cos^2 theta + 7 cos theta + 3 = 0

Now let u = cos theta getting
2u^2+7u=3=0 and factor as (u+3)(2u+1) = 0 -> u = -3 or u = -1/2
substitute back in for u so cos theta = -3 or cos theta = -1/2
cos theta is always between -1 and 1 so it can't be -3, so exclude that possibility
but for the cos theta = -1/2 ask yourself what angle is the cos of the angle equal to 1/2? and you will realize this happens when theta is pi/3 but you need to now analyze what this is suggesting the in the interval -720 deg to 0 degrees which is the same as saying [-4pi,0]. You will want the angle in the quadrants where cos theta is negative or where x is negative which is in the 2nd and 3rd quadrants. Figure out what angles are the reference angles. It just so happens that at 2pi3 the cos of 2pi/3 is -1/2 (2nd quadrant) and at 4pi/3 cos 4pi/3 = -1/2 (3rd quadrant). In the interval
[-4pi,0] this is found like this: Since the angle is parts of 1/3 pi take 2pi and convert it to
(6/3) pi. Now take from 2pi/3 two pi: 2pi/3 - 6pi/3 = -4pi/3 and do the same for 4pi/3
4pi/3-6pi/3 = -2pi/3. You have found two angles in the interval [-2pi,0]. To find the remaining
angles in [-4pi,-2pi] subtract another 2 pi from the two you just found: -4pi/3-6pi/3 = -10pi/3 and
-2pi/3-6pi/3 = -8pi/3. So the answer to the question is:
cos (-2pi/3) = cos(-4pi/3) = cos(-8pi/3) = cos(-10pi/3) = -1/2

If you are not comfortable in calling the angles pi, remember that 2pi = 360 deg so pi = 180 deg and everywhere you see pi put int 180 deg:
-4pi/3 = -4(180)/3 = -240 deg, -2pi/3 = -2(180)/3 = -120 deg, -8pi/3 =-8(180)/3 = -480 deg,
and finally -10pi/3 = -10(180)/3 = -600 deg and there you are.

Math Prof



---------- FOLLOW-UP ----------

QUESTION: 1. Prove that cotA = cosec2A+cot2A

2. Hence, by using the above identity or otherwise, show that cot 15 degree = 2+square root of 3

Answer
J.A.~
Start with one side or the other and try to arrive with the opposite side. I will begin with the 'more' complicated side:

csc 2A + cot 2A = 1/sin 2A + cos 2A/sin 2A = (1+ cos 2A)/sin 2A = (1 + 2cos^2 A - 1)/(2sin A cos A)
= 2cos^2 A/2sin A cos A = cos A/sin A = cot A

cot 15 deg = csc 2(15) deg + cot 2(15) deg = csc 30 deg + cot 30 deg
= 1/sin 30 deg + cos 30 deg/sin 30 deg = 1/(1/2) + [(sqrt3)/2]/(1/2) = 2 + sqrt 3

Math Prof