Express each of the following in the form r sin ( x + α ), where r>0 and 0<α<2pi 1. `sin x +sqrt3 cos x = 2 sin (x + pi/3)` Steps to solving this, I'm having troubles with the...

Express each of the following in the form

r sin ( x + α ), where r>0 and 0<α<2pi

1. `sin x +sqrt3 cos x = 2 sin (x + pi/3)`

Steps to solving this, I'm having troubles with the transformation steps

I have a quiz on this tomorrow, thanks in advance!

I really appreciate your help enotes teachers!!

Asked on by lederpina

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txmedteach's profile pic

txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

We will start with the part on the left:

`sinx + sqrt3cosx`

Now, there are two trig identities that will give us a sum of sines and cosines:

`sin(alpha+beta) = sinalphacosbeta + cosalphasinbeta`

`cos(alpha-beta) = cosalphacosbeta + sinalphasinbeta`

Let's just say that `alpha = x` so we'll be looking for `beta` to get things in the right form. It looks like we're just looking for the sin formula, so we can ignore the cos identity above.

However, to get the sine formula, we need to change the 1 in front of sin(x) to a cosine, and the square root of 3 in front of the cos(x) to a sin. This will be a problem!

Remember, `sqrt(sin^2beta + cos^2beta) = 1` by the pythagorean theorem on the unit circle. We can use this to see what we'd need to factor out of the expression so we can continue with our problem. Let me show you how (c is going to be the factor we take out):

`1 = 1/c sqrt(1^2 + sqrt(3)^2) = 1/c sqrt4 = 2/c`

So, in order for `2/c=1`, c must be 2. Therefore, we should be able to continue after we factor out a 2:

`sinx + sqrt3cosx = 2(1/2sinx + sqrt3/2 cosx)`

Well, that looks great! I only say this, of course, because now we have two numbers that look like a sine and cosine! Now, what we're trying to figure out is the angle `beta` such that:

`cosbeta = 1/2`

`sinbeta = sqrt3/2`

Remember, we know this because it has to fit with our `sin(alpha+beta)` formula from before.

Now, we can reason this out pretty easily. Both values are positive, so `beta` must be between 0 and `pi/2`. Now, you only get `sqrt3/2` and `1/2` when we're dealing with angles of `pi/6` or `pi/3`, and because our sine is the larger value, we can now say safely that `beta = pi/3`.

So our equation becomes:

`2(1/2sinx + sqrt3/2cosx) = 2(cos(pi/3)sinx + sin(pi/3)cosx)`

Now the inside of the parentheses fits exactly with our `sin(alpha+beta)` formula. Remember, for this case, `alpha =x` and `beta = pi/3`, giving us our final answer:

`2(cos(pi/3)sinx + sin(pi/3)cosx) = 2sin(x + pi/3)`

And you're done! I hope that worked for you. If you want more practice, show how you can get the same original equation into a form that looks like `2cos(x-beta)` (`beta` will be different for this one...).

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