What are the maximum/minimum values of 2sinx - 3cosx?

Question

Using one of the trig identities you can write this as in the form Rsin(x - a) = (√13)cos(x - 0.983). I've checked this part of this part of my answer and that's right...what I don't get is the next part...

We're asked to now write the min/max values of 1 + 2sinx - 3cosx (i.e. 1 + original expression). By looking at the form Rsin(x - a), you can see the min value would be 1 - √13 and the max value as 1 + √13. If you look at the actual equation though (i.e. 1 + 2sin x - 3cosx), isn't the max value when x = Pi/2 (i.e. 1 + 2 - 0 = 3) and the min value when x = 0 (i.e. 1 + 0 - 3 = -2).

I was just wondering, why is my second method (i.e. solutions of 3 and - 2) wrong and the first method right? The original question is:

Express 2 sin θ − 3 cos θ in the form Rsin(θ − α), where R and α are constants to be determined, and
0 < α < Pi/2

Hence write down the greatest and least possible values of 1 + 2 sin θ − 3 cos θ

Thanks!

Answer 1

i) For all real values, cos function lies in [-1, 1]

So (√13)*cos(θ − α) lies in [-√13, √13]



ii) As of the above, 1 + {2*sin(θ) − 3*cos(θ)} lies in 1 + [-√13, √13]



So minimum value is (1 - √13) and

Maximum value is (1 + √13)

Answer 2

1 + 2 sin θ − 3 cos θ is a translation of 2sinx - 3cosx parallel to the y-axis by the vector (0, 1), (Sorry I cant write column vectors)



So your max for 1 + 2 sin θ − 3 cos θ will be (1 + √13.) + 1 and min will be (1 - √13) - 1

Answer 3

The 2nd method is wrong because t = pi/2 doesn't produce the maximum value. Why you'd assume that is beyond me.



http://www.wolframalpha.com/input/?i=y+%3D+1+%2B+2sin%28t%29+-+3cos%28t%29



y = 1 + 2sin(t) - 3cos(t)

dy/dt = 2cos(t) + 3sin(t)

dy/dt = 0

0 = 2cos(t) + 3sin(t)

-2cos(t) = 3sin(t)

-2/3 = tan(t)

t = arctan(-2/3)

t = 2.5535900500422256872170323026544 + pi * k

t = 2.55359 + pi * k produces your extreme values.

Answer 4

Just plug it in you graphic calculator, if you can't . 3 square + 2 Square = 9+4 =13

Answer 5

2sinx - 3cosx

= sqrt(2^2+3^2) ( 2sinx - 3cosx) / sqrt(2^2+3^2)

= sqrt(13) ( 2sinx/sqrt(13) - 3cosx/sqrt(13) )



Let cos t = 2/sqrt(13)

then sin t = 3/sqrt(13)



2sinx - 3cosx

= sqrt(13) ( 2sinx/sqrt(13) - 3cosx/sqrt(13) )

= sqrt(13) ( cos t * sin x - sin t * cos x)

= sqrt(13) sin( x -t)



Now -1 <= sin(x - t) <= 1



Therefore the max and min of

2sinx - 3cosx

is +- sqrt(13)