Solve the given problem that involves Trigonometry

[chwala]

Homework Statement

See attached.

Relevant Equations

Trigonometry

For part (a),

We know that $\cos (-θ)=\cos (θ)$ and $\sin (-θ)=-\sin (θ)$

$\cos (A-B)=\cos A\cos (-B) -\sin A\sin(-B)$

$\cos (A-B)=\cos A\cos (B) +\sin A\sin(B)$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

For part (b)

...

$f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)$
$f(θ)=\cos 2θ+ \sin (θ+30^0)\sin (θ-30^0)$

on addition,
...
$2f(θ)= \cos 60^0 + cos 2θ$

$2f(θ)=\dfrac{1}{2}+ (2\cos^2 θ -1)$

$f(θ)=\cos^2 θ - \dfrac{1}{4}$For part (c) i,

The maximum value of $f(θ)$ is given by,

$f(180^0)=\left[1-\dfrac{1}{4}\right]=\dfrac{3}{4}$ at smallest positive value of $θ=180^0$

The minimum value of $f(θ)$ is given by,

$f(90^0)=\left[0-\dfrac{1}{4}\right]=-\dfrac{1}{4}$ at smallest positive value of $θ=90^0$

Bingo!

Any insight/alternative approach is highly welcome. Cheers guys.

 

[SammyS]

What you have for part (b) is so wrong that it's difficult to make any guess as to what you were trying to do.

Clearly it can't be that both of the following statements are true:

For part (b)

$f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)$
$f(θ)=\cos 60^0+ \sin (θ+30^0)\sin (θ-30^0)$

on addition,
...

Furthermore, on addition they give:

$2f(θ)=2\cos 60^\circ$

 

[chwala]

What you have for part (b) is so wrong that it's difficult to make any guess as to what you were trying to do.

Clearly it can't be that both of the following statements are true:Furthermore, on addition they give:

$2f(θ)=2\cos 60^\circ$

I just amended.

...small typo boss! That should be clear to you now.

 

[SammyS]

I just amended.

...small typo boss! That should be clear to you now.

That does give some small hint as to what you were trying to do. The only thing clear about this is that the result does not follow from what is given.

Suggestion:

Use part (a) to get a general result for $\cos(A+B) \cos(A-B)$. You get a difference times a sum, which factors to a difference of squares. Use that to get $f(\theta)$ plugging in $\theta$ for $A$ and $30^\circ$ for $B$ .

 

[chwala]

That does give some small hint as to what you were trying to do. The only thing clear about this is that the result does not follow from what is given.

Suggestion:

Use part (a) to get a general result for $\cos(A+B) \cos(A-B)$. You get a difference times a sum, which factors to a difference of squares. Use that to get $f(\theta)$ plugging in $\theta$ for $A$ and $30^\circ$ for $B$ .

...this should be simple sir, I just made use of part (a);
Specifically, $\cos(A+B)$ and $\cos (A-B)$...that gave me the two simultaneous equations.

 

[SammyS]

...this should be simple sir, I just made use of part (a);
Specifically, $\cos(A+B)$ and $\cos (A-B)$...that gave me the two simultaneous equations.

That may be what you intended to do, but you totally messed it up.

 

[chwala]

That may be what you intended to do, but you totally messed it up.

fine, i hope you're having a great day...let me just repeat what i did.Let

$(θ+30^0)=A$

and

let $(θ-30^0)=B$

Then,

$\cos(A+B)=f(θ)-\sin A\sin B$
$\cos (A-B)=f(θ)+\sin A\sin B$

$\cos 2θ=f(θ)-\sin A\sin B$
$\cos 60^0=f(θ)+\sin A\sin B$

...

cheers boss!

 

[SammyS]

fine, i hope you're having a great day...let me just repeat what i did.Let

$(θ+30^0)=A$

and

let $(θ-30^0)=B$

Then,

$\cos(A+B)=f(θ)-\sin A\sin B$
$\cos (A-B)=f(θ)+\sin A\sin B$

$\cos 2θ=f(θ)-\sin A\sin B$
$\cos 60^0=f(θ)+\sin A\sin B$

...

cheers boss!

OK. That does appear to work. In my opinion, unless you give that sort of explanation, it's next to impossible to see that your solution is correct.

I think that the more straight forward way to get the result would be to let $A=\theta$ and $B=30^\circ$, then proceed:

$\displaystyle \cos(A+B) \cos(A-B)=\left\{ \cos A \cos B - \sin A \sin B \right \} \left\{ \cos A \cos B + \sin A \sin B \right \}$

$\displaystyle = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B$

$\displaystyle = \cos^2 \theta \cos^2 30^\circ - \sin^2 \theta \sin^2 30^\circ$

etc.