Question 1167173: Question: TanA = v; A is in Quadrant 1. Obtain Sin2A and Cos2A.
My work so far: Assume the hypotenuse as 1, designate the sides X and Y (corresponding to axis). And I can get appropriate ansers based on the formula in abstract form.
But I have no idea as to how to approach the problem to come to answers given in the book as Sin2A = 2v/(1+v^2) and Cos2A = (1 - v^2)/(1+v^2)
Do I disignae Hypotenuse as v and the sides 1 and 2?
Or start working on Sin= v*Cos and Cos= Sin/v ?
Please help. Or if this question is already solved and posted on this site, just give me the question number and I can look it up.
Thanks
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
I think you are to draw an isosceles triangle and use the law of sines and the
law of cosines on it.
Since A is in Quadrant I, it is acute. So we draw an isosceles triangle with
each base angle equal to A, and the perpendicular bisector of the base, which
we let be v, and each half of the base be 1. Then the vertex angle (angle at
the top) will be 180°-2A and the base equal to 2, like this. Then, since the
isosceles triangle is split into two congruent right triangles, by the
Pythagorean theorem, each leg of the isosceles triangle is √(1+v2)
By the law of sines, the ratio of the length of a side of a triangle to the
sine of the angle opposite that side is the same for all sides and angles in
any triangle, so for the whole isosceles triangle with base 2,
 . That's because the sine of an angle is equal to
the sine of its supplement. So,
, and therefore
Cross-multiply,
From the right triangle on the left, sin(A)=opposite/hypotenuse,
Multiply both sides by √(1+v^2)
Divide both sides by 1+v2
--------------------------
Next we use the law of cosines with the same triangle above:
The square of a side of a triangle is equal to the sum of the squares of the
other two sides minus twice the product of the other two sides multiplied by
the cosine of the angle between them.
 . That's because the cosine of an angle is equal
to the negative of the cosine of its supplement. So we simplify and substitute
-cos(2A) for cos(180°-2A)
Divide through by 2
Solve for cos(2A)
Edwin
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