Trig questions!?Watch

#1
Hello just some questions on trigonometry:

For each of the following functions determine the maximum and minimum values of 'y' and the least positive values of 'x' at which these occur.

(a) y = 1 + cos2x

(b) y = 5 - 4sin(x + 30)

(c) y = 8 - 3cosÂ²x

Thanks
From John
0
12 years ago
#2
-1 =< cosA =< 1
-1 =< sinA =< 1

(a) y = 1 + cos2x
max value of cos2x = 1
so max value of y is 2.
min value of cos2x = -1
so min value of y is 0.

same for all other parts.
0
12 years ago
#3
or u can just differentiate... and take dy/dx = 0 ... which is more general,... but in this case... the first way is easier!
0
12 years ago
#4
differentiate.

a) dy/dx = 0 + -2sin2x = 0, x = 0, pi/2...
d^2y/dx^2 = -4cos2x, find the least one where it's >0

b) y = 5-4sin(x+30)
dy/dx = -4cos(x+30) = 0, x = pi/3, 4pi/3...
d^2y/dx^2 = 4sin(x+30)...

c) y = 8-3cos^2x = 8 = 3cosxcosx
dy/dx = 3(-2sinxcosx) = 3sin2x = 0, as above...
0
#5
hmmm i dont get it y = 1 + cos2x

How can u just rearrange it to 1= cos2x when there's a y it isnt "0" so in reality it would be

y-1=cos2x ?

Thanks
From John
0
12 years ago
#6
(Original post by Neo1)
hmmm i dont get it y = 1 + cos2x

How can u just rearrange it to 1= cos2x when there's a y it isnt "0" so in reality it would be
It's not a rearrangement: it's just saying that for y to be at a maximum, cos2x must be at a maximum. The maximum value of cos2x is 1, therefore the solution for x is the solution for cos2x = 1.
0
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