# Math Help - way to express ans

1. ## way to express ans

Can anyone help to check if my communication skill for part B is good enough please?

The question is as following:

Part a:
transform $2 \cos(\theta) + \sin(\theta)$ into the form r \sin (\theta + \alpha), where r > 0 and $0 \deg < \alpha < 90 \deg$.

>>>>> my ans obtained is
[tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

Part B
Given that $- \alpha \<= \theta \ <= 180\deg - \alpha$, find the value of k such that the equation

$2\cos \theta + 3 \sin \theta = k$
has a unique solution.

first I sketched (not draw) a graph of [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

then :
Given $- \alpha \<= \theta \ <= 180\deg - \alpha$
ie $- 33.7 \deg <= \theta \ <= 146.3 \deg$
From the sketched graph, it is only possible for k to have a unique solution, where $- \alpha \<= \theta \ <= 180\deg - \alpha$ when [tex]\sqrt{13} [/Math]

2. Originally Posted by ling_c_0202
Can anyone help to check if my communication skill for part B is good enough please?

The question is as following:

Part a:
transform $2 \cos(\theta) + \sin(\theta)$ into the form r \sin (\theta + \alpha), where r > 0 and $0 \deg < \alpha < 90 \deg$.

>>>>> my ans obtained is
[tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]
We want to rewrite

$
2 \cos(\theta) + \sin(\theta)=r \sin (\theta + \alpha)
$

So we may use the trig identity for the sin of the sum of two angles to
expand the RHS:

$
2 \cos(\theta) + \sin(\theta)=r (\sin (\theta) \cos( \alpha)+\cos(\theta) \sin(\alpha))
$

so we require:

$
2=r \sin(\alpha)
$

$
1=r \cos(\alpha)
$

Squaring and adding these equations give:

$
2^2+1^2=r^2 ((\cos (\alpha))^2 + (\sin(\alpha))^2)=r^2
$

so $r=\sqrt{5}$.

Also dividing the first equation by the second gives:

$
2=\tan(\alpha)
$

so $\alpha = \arctan(2) \approx 1.107$ radian, or $\approx 63.4 ^{\circ}$

RonL

3. Originally Posted by ling_c_0202
Can anyone help to check if my communication skill for part B is good enough please?

The question is as following:

Part a:
transform $2 \cos(\theta) + \sin(\theta)$ into the form r \sin (\theta + \alpha), where r > 0 and $0 \deg < \alpha < 90 \deg$.

>>>>> my ans obtained is
[tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]
Of course your solution corresponds to rewriting $2 \cos(\theta) +3 \sin(\theta)$ in the form $r \sin (\theta + \alpha)$.

RonL

4. Originally Posted by ling_c_0202
[Part B
Given that $- \alpha \<= \theta \ <= 180\deg - \alpha$, find the value of k such that the equation

$2\cos \theta + 3 \sin \theta = k$
has a unique solution.
We are interested in what values of $k$ result in a unique solution
to the equation:

$2\cos \theta + 3 \sin \theta=\sqrt{13} \sin(\theta+\alpha) = k$

for $- \alpha <= \theta <= 180\deg - \alpha$.

Put $\phi=\theta+\alpha$, then the question becomes: For which
values of $k$ does the equation:

$
\sqrt{13} \sin(\phi) = k
$

have a unique solution for $0 <= \phi <= 180\deg$.

The solution is:

$
\phi=\arcsin(k/\sqrt{13})
$
,

which is unique for $\phi$ in the specified range when:

$
k/ \sqrt{13}= 1
$
,

or:

$
k= \sqrt{13}
$
,

all other values of k give two solutions in the range, this is illustrated in the
attached plot.

RonL

5. Originally Posted by ling_c_0202
Can anyone help to check if my communication skill for part B is good enough please?

The question is as following:

Part a:
transform $2 \cos(\theta) + \sin(\theta)$ into the form r \sin (\theta + \alpha), where r > 0 and $0 \deg < \alpha < 90 \deg$.

>>>>> my ans obtained is
[tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

Part B
Given that $- \alpha \<= \theta \ <= 180\deg - \alpha$, find the value of k such that the equation

$2\cos \theta + 3 \sin \theta = k$
has a unique solution.

first I sketched (not draw) a graph of [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

then :
Given $- \alpha \<= \theta \ <= 180\deg - \alpha$
ie $- 33.7 \deg <= \theta \ <= 146.3 \deg$
From the sketched graph, it is only possible for k to have a unique solution, where $- \alpha \<= \theta \ <= 180\deg - \alpha$ when [tex]\sqrt{13} [/Math]
If you know how to graph or sketch sine and cosine curves then here is one way to do your two problems here.

1.) Express [2cos(theta) +sin(theta)] into [r*sin(theta +Alpha)], where Alpha is a constant and it is between 0 and 90deg, and r is greater than zero.

You graph y = 2cos(theta) and y = sin(theta) on the same (y,theta) cartesian axes.
The sum of the two y's is the y of r*sin(theta +Alpha).

Alpha, by the way, is the horizontal shift. Let's call it angle A.

In those two graphs, you see that
---at theta=0, r*sin(theta +A) = 2.
---at theta=90deg, r*sin(theta +A) = 1
---at theta=180deg, r*sin(theta +A) = -2.
---Etc.

So we use two of those.

When theta=0,
r*sin(0+A) = 2
sin(A) = 2/r -----(1)

When theta=90deg,
r*sin(90+A) = 1
r*[sin(90deg)cosA +cos(90deg)sin(A) = 1
r*[1*cosA +0*sinA] = 1
cosA = 1/r -----(2)

(1) divided by (2),
sinA/cosA = (2/r)/(1/r)
tanA = 2
A = arctan(2) = 63.435 degrees ----***
Substitute that into, say, (2),
cos(63.435deg) = 1/r
r = 1/cos(63.435deg) = 2.236 -----***

Therefore, 2cos(theta) +sin(theta) = 2.236sin(theta +63.435), in degrees. ----answer.

----------------------
2.) Find k so that [2cos(theta) +3sin(theta) = k] has unique solution.
where theta is between -Alpha and (180 -Alpha)deg.

You found
2cos(theta) +3sin(theta) = sqrt(13)*sin(theta +33.7deg)
Which is correct.

So Alpha = 33.7deg'
In the graph of y = sqrt(13)*sin(theta +33.7deg), based from the graphs of y=2cos(theta) and y=3sin(theta), the horizontal shift is to the left of the origin (0,0).
Meaning,
---the "start"--where it is zero--of the resulting sine curve is at theta = -33.7 deg.
---so the "end"--where is goes down to zero again--of the curve that is above the theta-axis (horizontal axis) is at (180 -33.7) deg.

The positive half of the graph of a sine curve is in the first half of the first cycle or period.
One cycle of y = sqrt(13)*sin(theta +33.7deg) is 360 degrees long, so the first half of that is from 0 to 180deg.

Meaning ,further, that in the interval theta = [-33.7,(180-33.7)], the curve of [2cos(theta) +3sin(theta)] is above the theta- or horizontal axis.

As seen from y = sqrt(13)*sin(theta +33.7deg), the amplitude is sqrt(13). So the curve went from 0 at theta=(-33.7deg), went to a maximum of sqrt(13) as it goes to the right, then came down to zero again at theta=(180 -33.7)deg.

Therefore, for 2cos(theta) +3sin(theta) = k, k=[0,+sqrt(13)], or k must be from zero to positive sqrt(13), so that there will be unique solutions to the said equation in the interval theta = [-33.7deg,(180-33.7)deg]. ---------answer.

6. Originally Posted by ticbol

Therefore, for 2cos(theta) +3sin(theta) = k, k=[0,+sqrt(13)], or k must be from zero to positive sqrt(13), so that there will be unique solutions to the said equation in the interval theta = [-33.7deg,(180-33.7)deg]. ---------answer.
That any value of $k \in [0,\sqrt{13}]$ does not give a unique solution to:

$
2 \cos(\theta) +3\sin (\theta) = k
$
,

is illustrated by the attached plot. This shows that $k=1$ gives two
solutions in the range, this is generally the case except when $k=\sqrt{13}$,
when the two solutions merge to give a unique solution.

RonL

7. Originally Posted by CaptainBlack
That any value of $k \in [0,\sqrt{13}]$ does not give a unique solution to:

$
2 \cos(\theta) +3\sin (\theta) = k
$
,

is illustrated by the attached plot. This shows that $k=1$ gives two
solutions in the range, this is generally the case except when $k=\sqrt{13}$,
when the two solutions merge to give a unique solution.

RonL
Umm, yes, you are right. When k=1, there are two values of theta in that positive first half of the graph of y = 2cos(theta) +3sin(theta).
I missed that.

But k is not +,-sqrt(13) as you said before in your not-yet-edited answer then.

Oh, well, who does not make mistake.
I do. I do make mistakes always. Sometimes I could correct them before someone else points them.

That's why this is a forum.

8. Originally Posted by ticbol
Umm, yes, you are right. When k=1, there are two values of theta in that positive first half of the graph of y = 2cos(theta) +3sin(theta).
I missed that.

But k is not +,-sqrt(13) as you said before in your not-yet-edited answer then.

Oh, well, who does not make mistake.
I do. I do make mistakes always. Sometimes I could correct them before someone else points them.

That's why this is a forum.
I had misread the range for the solution, and was looking a full cycle of
the sine curve - just shows a picture always helps.

RonL

9. ## thank you

Thank you very much for your kindly help.

I have got it!! THZZ