Question 1203202
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I'll use x in place of theta
And use y in place of T


Here are two resources to check out
<a href="https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.05%3A_Derivatives_of_Trigonometric_Functions">https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.05%3A_Derivatives_of_Trigonometric_Functions</a>
and
<a href="https://math.libretexts.org/Courses/Mount_Royal_University/MATH_1200%3A_Calculus_for_Scientists_I/1%3A_Limit__and_Continuity_of_Functions/1.7%3A_Limit_of__Trigonometric_functions">https://math.libretexts.org/Courses/Mount_Royal_University/MATH_1200%3A_Calculus_for_Scientists_I/1%3A_Limit__and_Continuity_of_Functions/1.7%3A_Limit_of__Trigonometric_functions</a>


The first link talks about the derivative of trig functions. 
What we're interested in mainly is that
derivative of sine = cosine
or in terms of symbols
d/dx[ sin(x) ] = cos(x)


Rephrased another way
y = sin(x) leads to dy/dx = cos(x)


The second link is a supplement to explain two important trig limit identities.


If y = (2V/g)*sin(x) then dy/dx = (2V/g)*cos(x)
The (2V/g) portion is a constant.


Plug in x = pi/6 to get:
dy/dx = (2V/g)*cos(x)
dy/dx = (2V/g)*cos(pi/6)
dy/dx = (2V/g)*sqrt(3)/2
dy/dx = (V/g)*sqrt(3)


Use the unit circle to determine that cos(pi/6) = sqrt(3)/2


Answer: The instantaneous rate of change is (V/g)*sqrt(3) when theta = pi/6


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Edit: 


I just realized you wrote that you haven't learned derivatives just yet.
Another approach would be to find the average rate of change over the interval [pi/6, pi/4] as done so in this link
<a href="https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1203157.html">https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1203157.html</a>


pi/6 = 0.523599 approximately
pi/4 = 0.785398 approximately


So the interval [pi/6, pi/4] is approximately [0.523599, 0.785398]


Then shrink that interval down to something like [0.523599, 0.7] then to [0.523599, 0.6]
The idea is to keep shrinking the interval such that pi/6 = 0.523599 is kept constant as the left endpoint. Only the right endpoint will move closer and closer to pi/6


As this right side approaches pi/6, the average rate of change will approach the instantaneous rate of change.
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