Thread: trigo again

Find the minimum value of the following expression and the value of x between 0 and 360 when the minimum occurs .

3 cos 2x + sin 2x --- this is the expression


3 cos 2x + sin 2x = r cos(2x-a)



tan a=1/3 , a =18.43

When minimum occurs , \sqrt{10}cos(2x-18.43)=-\sqrt{10}

cos(2x-18.43)=-1

2x-18.43=180 or 2x-18.43=270

Then i get my 2 values of x from there .

But i am wrong . Where is my mistake >?

Originally Posted by thereddevils

Find the minimum value of the following expression and the value of x between 0 and 360 when the minimum occurs .

3 cos 2x + sin 2x = r cos(2x-a)



tan a=1/3 , a =18.43

When minimum occurs , \sqrt{10}cos(2x-18.43)=-\sqrt{10}

cos(2x-18.43)=-1

2x-18.43=180 or 2x-18.43=270

Then i get my 2 values of x from there .

But i am wrong . Where is my mistake >?

Which expression are you trying to optimize? You have posted an equation with two expressions, one on each side.

I have not looked at the first part of your work, but this caught my eye:

cos(2x-18.43)=-1

2x-18.43=180 or 2x-18.43=270

That should be 540°, not 270°. The minimum values of the cosine function occur at odd multiples of 180°.

Now, although we have the restriction , you have a 2x, which means that . That's why the 2nd number above should be 540°.


01

Originally Posted by apcalculus

Which expression are you trying to optimize? You have posted an equation with two expressions, one on each side.

edited

Originally Posted by yeongil

I have not looked at the first part of your work, but this caught my eye:

That should be 540°, not 270°. The minimum values of the cosine function occur at odd multiples of 180°.

Now, although we have the restriction , you have a 2x, which means that . That's why the 2nd number above should be 540°.


01

thanks .. i should hv noticed that ....