[SOLVED] Getting the wrong sign for cosx+sinx=rsin(x+alpha)

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May 14th 2009, 07:22 PM
tleave2000

[SOLVED] Getting the wrong sign for cosx+sinx=rsin(x+alpha)

The question asks me to solve $7\cos x +6\sin x = 2$

So I put it into the form $r\sin (x + \alpha)$ :

$7\cos x +6\sin x \equiv r\sin (x + \alpha) \equiv r\sin x\cos\alpha + r\cos x\sin\alpha$

Then I equate the coefficients:
of cos x: $7=r\sin\alpha$ [1]
of sinx: $6=r\cos\alpha$ [2]

Divide [1] by [2] to get:
$\frac{7}{6}=tan\alpha \Rightarrow \alpha = 49.4^\circ$

Square and add [1] and [2] to get:
$7^2+6^2=r^2(\sin^2\alpha+\cos^2\alpha) \Rightarrow r=\sqrt{85}$

$\therefore 7\cos x+6\sin x \equiv \sqrt{85}\sin(x+49.4^\circ)$

Except that last statement doesn't seem to be true, according to
the answers in my book and when I have a look at the two curves in a
graphing application. I tried $\sqrt{85}\sin(x-49.4^\circ)$ in the graphing app and
that seems to be correct...
but why??

What have I done wrong? Why am I getting a positive value for
alpha when I should be getting the same number but negative?

The only way I can see to get -49.4 would be by introducing a minus
sign when dividing [1] by [2]. But I don't see how to justify that.
Have I made a simple error? I can't spot it (Shake)...
Any help would be much appreciated, thanks in advance.
May 15th 2009, 02:45 AM
HallsofIvy

Your result, $\sqrt{85}sin(x+ 49.4)$ is correct. An easy way to check that is to take x= 0. $\sqrt{85} sin(49.4)$ = (9.219)(0.759)= 7 while obviously $\sqrt{85} sin(-49.4)$ = -7.
May 15th 2009, 03:07 AM
Soroban

Hello, tleave2000!

I see nothing wrong with your work.
Why are you convinced that alpha must be negative?

I was taught a different approach . . .

Quote:

Solve: . $7\cos x +6\sin x \:=\: 2$

Divide by: $\sqrt{7^2+6^2} \:=\:\sqrt{85}\!:\quad \frac{7}{\sqrt{85}}\cos x + \frac{6}{\sqrt{85}}\sin x \:=\:\frac{2}{\sqrt{85}}$

Let $\alpha$ be an angle such that: . $\sin\alpha = \tfrac{7}{\sqrt{85}},\;\cos\alpha = \tfrac{6}{\sqrt{85}}$

Then we have: . $\sin\alpha\cos x + \cos\alpha\sin x \:=\:\frac{2}{\sqrt{85}}$

. . which is equivalent to: . $\sin(x + \alpha) \:=\:\frac{2}{\sqrt{85}}$

And we have: . $x + \alpha \:=\:\arcsin\left(\tfrac{2}{\sqrt{85}}\right) \;\approx\;12.53^o$

. . Then: . $x \;=\;12.53^o - \alpha$

Since $\sin\alpha = \frac{7}{\sqrt{85}}$ , then: . $\alpha \:=\:\arcsin\left(\tfrac{7}{\sqrt{85}}\right) \:\approx\:49.40^o$

. . Therefore: . $x \;=\;12.53^o - 49.40^o \;=\;-36.87^o$
May 15th 2009, 07:50 AM
tleave2000

oops

Ah thanks. I was convinced it should be negative because when you give degree values to a graph plotting application that's expecting radian values for it's trig functions (Wondering), then when you compare the graphs of $7\cos x+6\sin x$ and $\sqrt{85}\sin(x+49.4^\circ)$ they aren't the same... but oddly if you stick a minus sign before the offset, the graphs match up. (Is that just a coincidence?)

So that's what I was doing wrong, I needed to convert the input by multiplying it by pi/180. Thank you both again, and it was interesting to see an alternative way to get to rsin(x+alpha).