Question

Maximum and minimum values of the expression \(y = \cos \theta \left(\sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha}\right)\)?

Answer 1

@ganeshie8 So I considered the general rule of thumb, i.e., converting the expression to \(a \cos t + b\sin t\).

Identities that come to my mind are \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\) and \(\cos \theta \sin \theta = \frac{1}{2} \sin 2\theta\).

Answer 2

Well, did you get any noticeable thing by applying the above identities?

Answer 3

I didn't...

Answer 4

But I'm pretty sure that I should.

Answer 5

Cool. I'll give it a try then.

Answer 6

My main concern is removing the stupid sqrt.

Answer 7

Or do I have to do that? Maybe I can manage with the sqrt.

Answer 8

I'm getting something similar to this :

\(\cfrac{1}{2} \sin 2 \theta + \cos \sqrt{1 - \cos (\theta + \alpha) \cos (\theta - \alpha) } \)

(After simplification)

Answer 9

same here

Answer 10

Hmm! What if we square both sides? Will that help us?

Answer 11

that may be a good idea

Answer 12

Hmm! It is getting more complicated if we square both sides, I guess. :/

Answer 13

Should we attempt for a Calculus method?

Answer 14

sure.

Answer 15

*

Answer 16

@ganeshie8 - Help please :D

Answer 17

looks like I am getting to something through the squaring thing

Answer 18

Cool!

Answer 19

Yup, getting a quadratic expression in \(\cos 2\theta.\)

Answer 20

Oh, great. Good work

Answer 21

\[y^2 = \dfrac{-2\cos^2 2\theta + 2\sin^2 \alpha \cos 2 \theta +( 2\sin^2\alpha + 1 )}{4}\]

Answer 22

The problem with squaring is that (a) it took me a while to find this expression and (b) even though squaring can help me with the max, it can't help me with the min.

Answer 23

wait is that expression even right? >_<

Answer 24

OK, simple fix: add 1 to the numerator.

Answer 25

Whoops, I meant subtract. Yes.

Too many mistakes... =(

Answer 26

\[y^2 = \dfrac{-2 \cos^2 2\theta + 2\sin^2\alpha \cos 2\theta + 2\sin^2 \alpha }{4}\]

Answer 27

Scratch that. I meant add.

Answer 28

what is wrong with me

Answer 29

Can someone please suggest an easier way to do this? :(

Answer 30

used vectors..