# More trig max and min valuesWatch

This discussion is closed.
#1
5 + 3 sin^2 x + 4 cos^2 x

How do you deal with the squared sin and cos?

I'm supposed, I think to use the R sin/R cos and maybe double angle formulas as that is as far as the course goes at this stage.

Heeelp....

jorge
0
14 years ago
#2
what does the solution equal to first?

ie. 5 + 3 sin^2 x + 4 cos^2 x = ?
0
14 years ago
#3
sin^2 x = 1 - cos^2 x

so 5 + 3 sin^2 x + 4 cos^2 x
= 5 + 3(1 - cos^2 x) + 4 cos^2 x
= 5 + 3 - 3 cos^2 x + 4 cos^2 x
= cos^2 x + 8

as cos^2 x must be positive, it's lowest value can be 0 when x=pi/2, 3pi/2, 5pi/2 etc.
therefore cos^2 x + 8 must at least equal 8.

the maximum value of cos^2 x is when x=0, pi, 2pi etc. which is 1. so the maximum value of the expression is 9.
0
14 years ago
#4
mik1w....why must cos^2 x be positive?
0
14 years ago
#5
(Original post by Leifde)
mik1w....why must cos^2 x be positive?
Just try and make it negative...

Ben
0
14 years ago
#6
(Original post by Leifde)
mik1w....why must cos^2 x be positive?
Because cos^2x = (Cosx)(Cosx)

A square will always be positive, regardless of the sign of Cosx.

Euclid.
0
14 years ago
#7
its even quicker just to think about the graphs - when x=pi/2 , sin^2 x = 1, but cos^2 x = 0, and when x = pi etc the opposite is true, so the graph y = 5 + 3 sin ^2 x + 4 cos^2 x oscillate between 8 and 9 as required.
0
#8
(Original post by Euclid)
Because cos^2x = (Cosx)(Cosx)

A square will always be positive, regardless of the sign of Cosx.

Euclid.
Thanks Euclid

The problem is as the exercise is in the Rcos/Rsin section, I thought there might be a way of solving it using Rcos/Rsin.

No?

Jorge
0
14 years ago
#9
(Original post by Jorge)
Thanks Euclid

The problem is as the exercise is in the Rcos/Rsin section, I thought there might be a way of solving it using Rcos/Rsin.

No?

Jorge
Use the double angle identity for sin^2(x) and cos^2(x). It will still give you the same result.

Euclid.
0
#10
(Original post by Euclid)
Use the double angle identity for sin^2(x) and cos^2(x). It will still give you the same result.

Euclid.

How do you do that?

You mean cos 2A = cos^2A - sin^2A ???

I'm still stuck.

Could you explain?

Thanks

Jorge
0
14 years ago
#11
(Original post by Jorge)
How do you do that?

You mean cos 2A = cos^2A - sin^2A ???

I'm still stuck.

Could you explain?

Thanks

Jorge
Sure, I meant

cos^2(x) = [1 + cos(2x)]/2
sin^2(x) = [1 - cos(2x)]/2

Euclid.
0
#12
(Original post by Euclid)
Sure, I meant

cos^2(x) = [1 + cos(2x)]/2
sin^2(x) = [1 - cos(2x)]/2

Euclid.

Sorry, to go on...

5 + 3 sin^2 x + 4 cos^2 x

5 + 3[(1-cos2x)/2] + 4[(cos2x +1)/2]

5 + 1.5 - 1.5 cos2x + 2 cos 2x + 4

5 + 5.5 - 1.5 cos2x + 2 cos 2x

10.5 - 3.5 cos2x

...and I can't use my Rsin / Rcos methods...

what am I missing?

Jorge
0
14 years ago
#13
(Original post by Jorge)
Sorry, to go on...

5 + 3 sin^2 x + 4 cos^2 x

5 + 3[(1-cos2x)/2] + 4[(cos2x +1)/2]

5 + 1.5 - 1.5 cos2x + 2 cos 2x + 4

Jorge
It's + 2

Euclid.
0
#14
(Original post by Euclid)
It's + 2

Euclid.

Thanks for the correction, but what I meant was that I have reduced everything to cis and therefore I cannot use the Rsin /Rcos method, or can I?

Jorge
0
14 years ago
#15
(Original post by Jorge)
Thanks for the correction, but what I meant was that I have reduced everything to cis and therefore I cannot use the Rsin /Rcos method, or can I?

Jorge
You can't use it because you have squared terms. If you have both squares, then you use the identity and solve easily (simple quadratic). If you have only a single square term, then still use the identity and solve a quadratic. When you have Asin(x) + Bcos(x) etc. then you resort the Rcos/Rsin etc. method. You don't want to make things harder than they have to be!

[I should make clear that 'the identity' I refer to is sin^2(x) + cos^2(x) = 1]

Ben
0
14 years ago
#16
(Original post by Jorge)
Thanks for the correction, but what I meant was that I have reduced everything to cis and therefore I cannot use the Rsin /Rcos method, or can I?

Jorge
What do you mean by the "Rsin /Rcos method"?

Expanding out the expression will give:

(17/2) + (1/2)cos(2x)

Now cos(2x) will be at most 1, and at least -1

So min{(17/2) + (1/2)cos(2x)} = 8
and max{(17/2) + (1/2)cos(2x)} = 9

Euclid.
0
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