Question

I'm very confused about the range of this..

Answer 1

\[\LARGE y=3|sinx|-2|cosx|\]

Answer 2

I know
\[ \LARGE 1=<|sinx|+|cosx|<= \sqrt{2}\]

Answer 3

y = 3|sinx| - 2|cosx| similar with
y = 3|sinx| - 2|sin(90-x)|

y_minimum when x = 0, it is y = 3(0) - 2(1) = -2
y_max when x = 90, it is y = 3(1) - 2(0) = 3

so, the range for y is
-2 <= y <= 3

Answer 4

why did you convert cosx to sin(90-x)
any specific reason?

Answer 5

@RadEn

Answer 6

that is an identity in trigono :)
cos(x) = sin(90-x)

Answer 7

i know that
i want to know why did you think of that?

Answer 8

in this question i mean any specific reason?

Answer 9

i trying to make y in sin terms, so that easier to get the output values

maybe :)

Answer 10

hmm

Answer 11

does anyone have any other methods ?just curious..thanks @RadEn though!

Answer 12

u can make y in cosine terms too
y = 3|cos(90-x)| - 2|cosx|

you're welcome :)

Answer 13

the cosine graph can be though as the sine graph offset by 90 degrees phase. so basically RadEn's method is to simplify the graph.
Of course,
\(y=3|\sin x| - 2|\cos x|\)
y is maximum when the y=3|sin x| is max and y=2|cos x| is min. note that there cannot be negative values. for min y ,it is vice versa.
the max value of 3sin x is 3(sinx =1), while the min value of 2 cosx is 0. so that gives 3
the min value 3sin x is 0(sinx =0), while the min value of 2 cosx is 2. so that gives -2
so, -2<=y<=3

Answer 14

thanks!

Answer 15

you're welcome :)