<p>I am doing some self-preparation for the Math II with the Barrons but I do not understand the following.</p>
<ol>
<li><p>Solve 2sinx + cos2x = 2sin^2 x -1 for x between 0 and 2pie
I understand that I have to enter the left side into Y1 and the right side into Y2. I know I
should set Xmin =0 and Xmin= 2pie
But could someone please explain to me why I Ymin=-4 and Ymax = 4</p></li>
<li><p>How do you find the range of the function f(x) = 5-6sin(pie x +1)
I only know that for a sin function, the range is between -1 and 1. but the book says the answer is x is between -6 and 6.</p></li>
</ol>
<p>1) I don’t know the calculator way, but here’s the trig way.
(cos2x= 1-2sin^x).
ques- 2sinx+cos2x=2sin^2 x -1
ans- 2sinx+1-2sin^2x=2sin^2x-1
4sin^2x-2sinx-2=0
2sin^2x-sinx-1=0
Facorise,
get sinx=-1/2 and sinx=1</p>
<p>Since the interval is [0,pie], we reject the negave value.Ans=1</p>
<p>2) Again, I don’t use a graphing calculator, but here’s the concept.
Range of sin=[-1,1]
f(x)=5-6sin(something).
You need the range of the function,so don’t think about the values of x. The minimim value of sin=(-1).
Put it in the eqn.
f(x)=5-6(-1)
=5+6=11. -> This is the max value of the fn.</p>
<p>The max value of x=1
Put in the eqn
f(x)= 5 -6(1)
= (-1). -> this is the min value of the fn.</p>
<p>The range could not be [-6,6]. Range means the possible values we can get of the function. For the range to be -6, the sin value would have to be 1.83 which is not possible.</p>