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Grade 11Trigonometry

Find the range of y.
Please provide the answer.

Question image for Find the range of y. Please provide the answer.

Shivansh Sangwan

8 Years agoGrade 11

4 Answers

Yogesh Kumar

8 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

8 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

8 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

8 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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