combine sine and cosine

barryb52 · Jan 23rd 2008, 09:13 AM

Hi folks been really stuck on this question on my homework, any help what so ever would be extremely apprieciated.

Write 3sin(x)-4cos(x) in the form Asin(x+x*) * l.h.s sign is an x but the left sides on it are joined not sure what the signs called lol* and give the amplitude and phase. Hence find the general solution of 3sin(x)-4cos(x)=2.

many thanks,
barry

**TKHunny** · Jan 23rd 2008, 10:26 AM

This is GREAT stuff. You do not want to be bad at it. Reread the section and make sure you get it. It will help you.

$3\sin(x)-4\cos(x)\;=\;$

$\sqrt{3^{2}+4^{2}}\;=\;5$

$\phi\;=\;atan(4/3)$

$3\sin(x)-4\cos(x)\;=\;5\sin\left(x - atan(4/3)\right)$

Can you answer the questions from there? If not, go back a couple of sections and read them over. You DO need this stuff.

**topsquark** · Jan 23rd 2008, 10:26 AM

Originally Posted by barryb52

Hi folks been really stuck on this question on my homework, any help what so ever would be extremely apprieciated.

Write 3sin(x)-4cos(x) in the form Asin(x+x*) * l.h.s sign is an x but the left sides on it are joined not sure what the signs called lol* and give the amplitude and phase. Hence find the general solution of 3sin(x)-4cos(x)=2.

many thanks,
barry

It would help to note that
$a~sin(x) + b~cos(x) = \sqrt{a^2 + b^2}~sin(x + \phi )$
where the phase angle
$\phi = sin^{-1} \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )$

Please don't ask me to derive it.

-Dan

**theanonomous** · May 3rd 2010, 10:09 PM

Although the formula $\phi = sin^{-1} \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )$ is correct, it might not give you the right answer when you enter this into your calculator; it is important to remember that this formula comes from the angle that a point (x=a, y=b) makes with the origin. This leads to the above formula and the equivalent formulas $tan^{\phi}=\frac{b}{a}$ and $<br /> \phi= cos^{-1} \left ( \frac{a}{\sqrt{a^2 + b^2}} \right )<br />$ and the quadrant $\phi$ is in is determined by the sine (+ or -) of a and b. The actual value of $\phi$ might be different from what your calculator tells you if $\phi$ is not in the quadrant your calculator is expecting it to be in. For example, if your a and b were negative your $\phi$ would be based on a point in the third quadrant and would therefore be between pi and 3pi/2 (in degrees, between 180 and 270). However, if you used the tangent formula and entered a negative a divided by a negative b the calculator would read this as a positive and number and give an answer in the 1s quadrant (between zero and pi/2, in degrees between 0 and 90). If you used the sin function your answer would be in the fourth quadrant, probably written as a negative value (between zero and -pi/2, in degrees between 0 and -90). Using the cos formula would give an answer in the 2nd quadrant (between pi/2 and pi, in degrees between 90 and 180). However, the actual value of $\phi$ is none of these. Therefore, the formula given in the previous would be more correctly written as $\phi = n\pi - sin^{-1} \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )$ where n=0 if a is positive and n=1 if a is negative. The cos formula would be better written as $\phi = n\pi - cos^{-1} \left ( \frac{a}{\sqrt{a^2 + b^2}} \right )$ where n=0 if b is positive and n=2 if b is negative. The tan formula would be better written as $\phi = tan^{-1} \left( \frac{b}{a} \right) +n\pi$ where n=0 is a is positive and n=1 if a is negative.

Basically, when solving a problem like the one presented, chose one of the three formulas to find a value for $\phi$ using your calculator, and check to make sure this answer is in the same quadrant as (b,a). If it isn't, and you were using the tan formula, add pi (180 degrees) to your answer. If $\phi$ is in the wrong quadrant and you were using the cos formula, reflect your answer around the x axis, and if you get an answer in the wrong quadrant using the sin formula reflect your answer about the y axis.

**harish21** · May 3rd 2010, 10:24 PM

Originally Posted by theanonomous

Although the formula $\phi = sin^{-1} \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )$ is correct, it might not give you the right answer when you enter this into your calculator; it is important to remember that this formula comes from the angle that a point (x=b, y=a) makes with the origin. This leads to the above formula and the equivalent formulas $tan^{\phi}=\frac{b}{a}$ and $<br /> \phi= cos^{-1} \left ( \frac{a}{\sqrt{a^2 + b^2}} \right )<br />$ and the quadrant $\phi$ is in is determined by the sine (+ or -) of a and b. The actual value of $\phi$ might be different from what your calculator tells you if $\phi$ is not in the quadrant your calculator is expecting it to be in. For example, if your a and b were negative your $\phi$ would be based on a point in the third quadrant and would therefore be between pi and 3pi/2 (in degrees, between 180 and 270). However, if you used the tangent formula and entered a negative a divided by a negative b the calculator would read this as a positive and number and give an answer in the 1s quadrant (between zero and pi/2, in degrees between 0 and 90). If you used the sin function your answer would be in the fourth quadrant, probably written as a negative value (between zero and -pi/2, in degrees between 0 and -90). Using the cos formula would give an answer in the 2nd quadrant (between pi/2 and pi, in degrees between 90 and 180). However, the actual value of $\phi$ is none of these. Therefore, the formula given in the previous would be more correctly written as $\phi = sin^{-1} \left ( \frac{b}{\sqrt{a^2 + b^2}} \right )+ \frac{n\pi}{2}$ where n=1, 2, or 3 and is determined by the quadrant that the point (b,a) lies in.

Basically, when solving a problem like the one presented, chose one of the three formulas to find a value for $\phi$ using your calculator, and check to make sure this answer is in the same quadrant as (b,a). If it isn't, add ether pi/2, pi, or 3pi/2 to your answer so that it is in the correct quadrant.

Hey the anonomous,

just out of curiosity, what made you comment on a post that is more than 2 years old? How did you even find it?

**theanonomous** · May 22nd 2010, 11:27 PM

LOL, at the time I found this post, I was trying to figure the answer to the question I answered on the post for a Differential equations take-home quiz. Even though the post didn't answer my question I figured it out by thinking about it and reviewing my notes, so I figured I'd be nice and putt a better answer for future frantic google searchers stuck in my position. Also that was my first post, and I also didn't realize how long it takes to takes to use that equation editor

, but once I got started I couldn't let all my hard work go to waste

.

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