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v1 = 0.02t + 1
v1 is the first part of the graph. So we simply need it's equation.Thnx for replying...........but I dont get the v1, how did u find it?
Sorry for the late reply. Could you explain how did you get the exact value of the inverse of cos 0.5?(i) f(x) = 4, so: 6 - 4cos(0.5x) = 4
cos(0.5x) = 0.5
0.5x = π / 3 (since it says exact, no need for basic angles and stuff)
x = 2π / 3
(ii) Max = a + b = 6+4 = 10 and Min = a - b = 6 - 4 = 2
So, range: 2 <f(x)<10
(iii) The graph is going to be upside down since the cos is -ve.............. so it would be something like this: (sorry for the crappy drawing though)
View attachment 57098
(iv) The inverse is just like the normal one
Hope it helps...............
These are the common angles and their trigonometric ratios you should try to learn:
But thats the problem. I dont memorize these values but instead I calculate them, therefore I am unable to get the values of their inverse.These are the common angles and their trigonometric ratios you should try to learn:
As you can see, cos^-1(0.5) = 60°, or π/3 radians.
Yesssss! Finally got a way to calculate these exact values!! thanks though qwertypoiu!These are the common angles and their trigonometric ratios you should try to learn:
As you can see, cos^-1(0.5) = 60°, or π/3 radians.
Yeah I understood it. Thanks a lot for your help!(i) f(x) = 4, so: 6 - 4cos(0.5x) = 4
cos(0.5x) = 0.5
0.5x = π / 3 (since it says exact, no need for basic angles and stuff)
x = 2π / 3
(ii) Max = a + b = 6+4 = 10 and Min = a - b = 6 - 4 = 2
So, range: 2 <f(x)<10
(iii) The graph is going to be upside down since the cos is -ve.............. so it would be something like this: (sorry for the crappy drawing though)
View attachment 57098
(iv) The inverse is just like the normal one
Hope it helps...............
year??
f(1) = -1/2 ?
Welcome................sorry didnt reply to ur msgs before (I just logged in now)..............glad i helped........
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