Question:
what is the trig identity for cos (x+ h) =?
Rae
2010-07-05 12:32:18 UTC
i know its not actually x+h but i cant draw the little triangle x thing on the computer. please help or ask questions if you dont understand what im asking.
Three answers:
Raymond
2010-07-05 12:40:26 UTC
Do you mean

cos(x+∆h)



If so, then



cos(x+∆h) = cos(x)cos(∆h) - sin(x)sin(∆h)



This symbol ∆ is normally used to indicate a "small amount of" or a difference.



It is used when figuring out where derivative come from. Since the derivative is the instantaneous slope, you try to find how fast a function grows, when x is increased by a small amount (for example, by ∆h).



The increase is the difference



cos(x+∆h) - cos(x)



And the slope is



[cos(x+∆h) - cos(x)] / ∆h



The trick is to find the limit as ∆h gets smaller and smaller.



---



As for the limit itself, it is found in this way:



[cos(x+∆h) - cos(x)] / ∆h

can be rewritten:

[cos(x)cos(∆h) - sin(x)sin(∆h) - cos(x)] / ∆h



rearrange the order:

[cos(x)cos(∆h) - cos(x) - sin(x)sin(∆h)] / ∆h



Split the numerator:to separate the cos from the sin:

[cos(x)cos(∆h) - cos(x)] / ∆h - [sin(x)sin(∆h)] / ∆h



The first term:

we know cos(0) = 1

as ∆h approaches 0, cos(∆h)=1, leaving us with

cos(x)*1 - cos(x) = cos(x) - cos(x) = 0



The first term is gone.



The second term:

as ∆h gets smaller and smaller, we find that sin(∆h) = ∆h

(for angles in radians)

This is a trick used in astronomy, where angles can be very small.

This means that the limit, as ∆h approaches 0, is

sin(∆h) / ∆h = 1



Leaving us with

-[sin(x)sin(∆h)]/∆h = -sin(x)*1 = -sin(x)
anonymous
2010-07-05 19:34:49 UTC
Use the cosine addition formula:

cos(A + B) = cos(A)cos(B) - sin(A)sin(B).



With A = x and B = ∆x, we have:

cos(x + ∆x) = cos(x)cos(∆x) - sin(x)sin(∆x).



I hope this helps!
?
2010-07-05 19:33:56 UTC
cos(x+h)= cos(x)cos(h) - sin(x)sin(h)


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
Loading...