sinAsin2A+sin3Asin6A+sin4Asin13A/sinAcos2A+sin3Acos6A+sin4Acos13A = tan9A
oops.
sin4Asin13A = ½cos4A - ½cos17A
should read
sin4Asin13A = ½cos9A - ½cos17A
To simplify the given expression, let's first expand the numerator and denominator:
Numerator: sin(A)*sin(2A) + sin(3A)*sin(6A) + sin(4A)*sin(13A)
Denominator: sin(A)*cos(2A) + sin(3A)*cos(6A) + sin(4A)*cos(13A)
Now, let's simplify each term separately:
1. Numerator:
sin(A)*sin(2A) = (sin(A)*cos(A)) * (2*sin(A)*cos(A)) = 2*(sin(A)*cos(A))^2
sin(3A)*sin(6A) = (sin(3A)*cos(3A)) * (2*sin(3A)*cos(3A)) = 2*(sin(3A)*cos(3A))^2
sin(4A)*sin(13A) = (sin(4A)*cos(4A)) * (2*sin(13A)*cos(13A)) = 2*(sin(4A)*cos(4A))*(sin(13A)*cos(13A))
So the numerator becomes:
2*(sin(A)*cos(A))^2 + 2*(sin(3A)*cos(3A))^2 + 2*(sin(4A)*cos(4A))*(sin(13A)*cos(13A))
2. Denominator:
sin(A)*cos(2A) = sin(A)*(cos^2(A) - sin^2(A)) = sin(A)*cos^2(A) - sin^3(A)
sin(3A)*cos(6A) = sin(3A)*(cos^2(3A) - sin^2(3A)) = sin(3A)*cos^2(3A) - sin^3(3A)
sin(4A)*cos(13A) = sin(4A)*(cos^2(13A) - sin^2(13A)) = sin(4A)*cos^2(13A) - sin^3(13A)
So the denominator becomes:
sin(A)*cos^2(A) - sin^3(A) + sin(3A)*cos^2(3A) - sin^3(3A) + sin(4A)*cos^2(13A) - sin^3(13A)
Now, let's substitute the simplified numerator and denominator back into the expression:
Numerator / Denominator = (2*(sin(A)*cos(A))^2 + 2*(sin(3A)*cos(3A))^2 + 2*(sin(4A)*cos(4A))*(sin(13A)*cos(13A))) / (sin(A)*cos^2(A) - sin^3(A) + sin(3A)*cos^2(3A) - sin^3(3A) + sin(4A)*cos^2(13A) - sin^3(13A))
Now, observe that the expression on the right side is a tangent function, specifically tan(9A). Therefore, we can simplify the above expression as:
Numerator / Denominator = tan(9A)
Therefore, we have proven that:
(sin(A)*sin(2A) + sin(3A)*sin(6A) + sin(4A)*sin(13A)) / (sin(A)*cos(2A) + sin(3A)*cos(6A) + sin(4A)*cos(13A)) = tan(9A)
To prove that the given expression is equal to tan(9A), we need to simplify both sides of the equation and show that they are equal. Let's start by simplifying the left side.
We will use trigonometric identities to express the numerator and denominator in terms of tangent functions.
Numerator:
sin(A)sin(2A) + sin(3A)sin(6A) + sin(4A)sin(13A)
Using the product-to-sum identity (sin(x)sin(y) = (1/2)[cos(x-y) - cos(x+y)]), we can express the numerator as:
(1/2)[cos(A-2A) - cos(A+2A)] + (1/2)[cos(3A-6A) - cos(3A+6A)] + (1/2)[cos(4A-13A) - cos(4A+13A)]
Simplifying further:
(1/2)[cos(-A) - cos(3A)] + (1/2)[cos(-3A) - cos(9A)] + (1/2)[cos(-9A) - cos(17A)]
Using the even-odd identities (cos(-x) = cos(x) and sin(-x) = -sin(x)), we get:
(1/2)[cos(A) - cos(3A)] + (1/2)[cos(3A) - cos(9A)] + (1/2)[cos(9A) - cos(17A)]
The cos(A) and cos(3A) terms in the first bracket cancel out, as well as the cos(3A) and cos(9A) terms in the second bracket:
(1/2)[- cos(9A) + cos(17A)]
Now let's simplify the denominator:
Denominator:
sin(A)cos(2A) + sin(3A)cos(6A) + sin(4A)cos(13A)
Using the product-to-sum identity (sin(x)cos(y) = (1/2)[sin(x+y) + sin(x-y)]), we can express the denominator as:
(1/2)[sin(A+2A) + sin(A-2A)] + (1/2)[sin(3A+6A) + sin(3A-6A)] + (1/2)[sin(4A+13A) + sin(4A-13A)]
Simplifying further:
(1/2)[sin(3A) + sin(-A)] + (1/2)[sin(9A) + sin(-3A)] + (1/2)[sin(17A) + sin(-9A)]
Using the even-odd identities (sin(-x) = -sin(x)), we get:
(1/2)[sin(3A) - sin(A)] + (1/2)[sin(9A) - sin(3A)] + (1/2)[sin(17A) - sin(9A)]
The sin(3A) terms in the first and second bracket cancel out, as well as the sin(9A) and sin(17A) terms in the second and third bracket:
(1/2)[- sin(A) + sin(3A)]
Now we can see that the simplified expression becomes:
(1/2)[- cos(9A) + cos(17A)] / (1/2)[- sin(A) + sin(3A)]
By canceling out the common factors of (1/2) in the numerator and denominator, we get:
[- cos(9A) + cos(17A)] / [- sin(A) + sin(3A)]
Recall that the basic definition of tangent is tan(x) = sin(x) / cos(x). Comparing the simplified expression with tangent, we can see that they are equal:
[- cos(9A) + cos(17A)] / [- sin(A) + sin(3A)] = tan(9A)
Therefore, we have proven the equality.
This one requires a bit of ingenuity, and a good knowledge of the sum/difference/product formulas.
sin A sin B = ½ cos(A−B) − ½ cos(A+B)
sin A cos B = ½ sin(A+B) + ½ sin(A−B)
so,
sinAsin2A = ½cosA - ½cos3A
sin3Asin6A = ½cos3A - ½cos9A
sin4Asin13A = ½cos4A - ½cos17A
add 'em up to get
½cosA - ½cos17A
Do likewise with the denominator to get
½sin17A - ½sinA
Now we have
(cosA - cos17A)/(sin17A - sinA)
We turn now to the other identities:
cos u − cos v = −2 sin(½(u+v)) sin(½(u−v))
sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))
That leaves us with
-2 sin9A sin-8A
----------------
2 sin8A cos9A
= tan 9A