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Find the range of y. Please provide the answer.

Find the range of y.
Please provide the answer.

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Grade:11

4 Answers

Yogesh Kumar
43 Points
6 years ago
y=cosx(sinx+cosx)
 =cosxsinx+cos2x
 =2cosxsinx/2+(cos2x+1)/2
  using 2sinxcosx=sin2x
=sin2x/2+cos2x/2+1/2
 
Now ½ is constant and using acos acos\Theta +bsin\Theta has max range at \sqrt{a^{2}+b^{2}} and minimum at 
Yogesh Kumar
43 Points
6 years ago
 
y=cosx(sinx+cosx)
 =cosxsinx+cos2x
 =2cosxsinx/2+(cos2x+1)/2
  using 2sinxcosx=sin2x
=sin2x/2+cos2x/2+1/2
 
Now ½ is constant and using acos acos\Theta +bsin\Theta has max range at \sqrt{a^{2}+b^{2}} and minimum at -\sqrt{a^{2}+b^{2}}.
we get 
y max=\sqrt{1/2^{2}+1/2^{2}}=\sqrt{1/2} + 1/2 and y min=-\sqrt{1/2^{2}+1/2^{2}}=-\sqrt{1/2} + 1/2
so this is the range of y from -\sqrt{1/2} + 1/2 to \sqrt{1/2} + 1/2
Hope you understood this.
Please message me if there is any problem in the solution
Soumendu Majumdar
159 Points
5 years ago
Dear Student,
y=cosx(sinx+cosx)
Now 
-\sqrt{a^2+b^2}\leq asin\theta + bcos\theta\leq \sqrt{a^2+b^2}
And y= 2cosxsinx/2 + 2cos^2x/2
= sin2x/2 + (1+cos2x)/2    since sin 2A = 2 sin  A cos A & cos 2A = 2cos^2A -1
So 
-\sqrt{1/2}\leq sin2x/2 + cos 2x/2\leq \sqrt{1/2}
Hence the range of y is
-\sqrt{1/2}+1/2 --> \sqrt{1/2} +1/2
Hope it helps!
 
Soumendu Majumdar
159 Points
5 years ago
If you have any problem or confusion in understanding the solution, don’t hesitate feel free to ask, I’ll be glad to help you!
Thanks!
with regards,
Soumendu

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