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wow, it feels great when you finally solve something which you work on for a long time thanks for your help!

than its simply cosXo if I multiply it with the limit of 1st one :1 than the limit of m approaches cosXo?

oooh if x-->Xo therefore is it (cos(2*Xo))/2?

Im getting a lost a little bit, is it actually approaches zero?

I see what you mean, that (2) is in radian and not degrees, thus the answer to the original question is (cos(Xo))/2 right?

But don't you need to take the cosine of two?

isnt it approaches 1?

I don't really understand what you mean, could you please explain it in details?

Im doing the (b) part of the exercise, which ask about what do they approach as X approaches Xo (their limits) 1)sin((X-Xo)/2)/((X-Xo)/2) 2)cos ((X+Xo)/2) 3)m(=sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2) Note: Xo an X are points on the cos,sin curve which is not bigger than pi/2 I was able to...

oohhh, i found the formula in my exercise book, it was really that standard thank you for your help, now I am doing the (b) part of the exercise about their limits May I ask for your help if i get stuck?

Do you mean sin(A-B) = sinAcosB - cosAsinB ?

Proving slope "m" of a secant connecting two points of the sine curve Homework Statement Write and expression for the slope m of the secant connecting the points Po(Xo,Yo) and P(X,Y) of the sine curve. Use the appropriate trigonometric identity to show that m= sin((X-Xo)/2)/((X-Xo)/2) * cos...

i got the answer for 2) (its about 19.2min), does anyone have some hint for the 1st one?

for 3) of course i drew a picture, i could only get as far as the slope+tower all together is 25.37m, maybe i misdrew the picture or just couldn't continue from the answer (25.37) i got and 2) i used your hint and calculated out the height/distance ration, i tried to substitue it into the sin...

Hey guys, We've been learning cos/sin rule since the 1st of school, and i really had difficulties with the following exercises, I've been working them yesterday for 2-3 hours and could only manage to do part of it... 1. IN Quadrilateral ABCD, AB=7 cm, BC= 8cm, CD=5cm and angle ABC=52...