lim delta x to 0 of
sin[(pi/6)+delta x]-(1/2)
all divided by delta x
hope i made that clear.
hint given: sin(theta+gamma)=sin theta cos gamma+cos theta sin gamma
sorry not sure where to start on this one
first of all, to make things easier to type, let's say
h = delta x
so your question becomes
lim [ sin(π/6 + h) - sin π/6 ]/h , as h ---> 0
This looks like you want the derivative of
sin x, when x = π/6
if f(x) = sinx
f'(x) = cosx
f'(π/6) = cos π/6 = √3/2
HOWEVER, it looks they actually want you to evaluate this by First Principles.
Using the hint they gave you, this will be a messy messy calculation.
Here is a page where they show those steps, they use d instead of h, and you will have to replace x with π/6
http://www.math.com/tables/derivatives/more/trig.htm
A more tradional method is to use
sin A − sin B = 2 sin ½ (A − B) cos ½ (A + B)
as seen in this video
http://www.youtube.com/watch?v=T0HPGrUXNzY
(hope you can understand the Indian accent, also close the ad at the beginning)
To solve the limit, we can use the given hint and the properties of limits.
Let's simplify the expression step by step:
1. Apply the given identity: sin(theta + gamma) = sin(theta)cos(gamma) + cos(theta)sin(gamma).
The expression becomes:
[sin(pi/6)cos(delta x) + cos(pi/6)sin(delta x)] - (1/2) / delta x
2. Rewrite sin(pi/6) and cos(pi/6) using the exact values:
[1/2 * cos(delta x) + sqrt(3)/2 * sin(delta x)] - (1/2) / delta x
3. Distribute the delta x:
(1/2 * cos(delta x) + sqrt(3)/2 * sin(delta x) - (1/2)) / delta x
4. Factor out 1/2 from the numerator:
(1/2 * [cos(delta x) - 1] + sqrt(3)/2 * sin(delta x)) / delta x
5. Observe that as delta x approaches 0, cos(delta x) approaches 1 and sin(delta x) approaches 0. Thus, we have:
(1/2 * [1 - 1] + sqrt(3)/2 * 0) / delta x
6. Simplify further:
(0 + 0) / delta x
0 / delta x
0
Therefore, the limit of the expression as delta x approaches 0 is 0.
To solve this problem, we can use the hint given. Let's break it down step by step.
First, let's substitute the given expression into the hint:
sin(theta + gamma) = sin(theta)cos(gamma) + cos(theta)sin(gamma)
In our case, theta = pi/6 and gamma = delta x. Substituting these values, we get:
sin[(pi/6) + delta x] = sin(pi/6)cos(delta x) + cos(pi/6)sin(delta x)
Next, let's subtract (1/2) from both sides of the equation:
sin[(pi/6) + delta x] - (1/2) = sin(pi/6)cos(delta x) + cos(pi/6)sin(delta x) - (1/2)
Now, we can rewrite this expression as:
sin[(pi/6) + delta x] - (1/2) = sin(pi/6)(cos(delta x) - 1) + cos(pi/6)sin(delta x)
Finally, divide both sides of the equation by delta x:
(sin[(pi/6) + delta x] - (1/2))/delta x = [sin(pi/6)(cos(delta x) - 1) + cos(pi/6)sin(delta x)]/delta x
To evaluate the limit of this expression as delta x approaches 0, we can take the limit of each term separately.
As delta x approaches 0, sin(delta x)/delta x approaches 1 and (cos(delta x) - 1)/delta x approaches 0.
Therefore, the final result is:
lim delta x to 0 of (sin[(pi/6) + delta x] - (1/2))/delta x = sin(pi/6)(0) + cos(pi/6)(1) = cos(pi/6)
Therefore, cos(pi/6) is the value of the given limit.