Attachment 29582

I remember doing a lot of trigonometric identities but I dont really see how a) is supposed to be reduced into Asin(x + c). Does anyone see how this might be done?

Thanks in advance...

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- Oct 25th 2013, 08:44 PMsepotoExpress in terms of Asin(x + c)
Attachment 29582

I remember doing a lot of trigonometric identities but I dont really see how a) is supposed to be reduced into Asin(x + c). Does anyone see how this might be done?

Thanks in advance... - Oct 25th 2013, 09:15 PMSlipEternalRe: Express in terms of Asin(x + c)
Use the sum of angles formula for sin: $\displaystyle \sin(x+c) = \sin(x)\cos(c) + \cos(x)\sin(c)$.

Next, match terms. You have $\displaystyle \sin(x) + \sqrt{3}\cos(x)$. The sin(x) obviously goes with sin (x), so cos(c) must have something to do with the coefficient of sin(x), which is 1. Then cos(x) obviously goes with cos(x), so sin(c) must have something to do with the coefficient of cos(x), which is $\displaystyle \sqrt{3}$. Assume that the 1 and the $\displaystyle \sqrt{3}$ are sides of a right triangle. What would the hypotenuse be? By the Pythagorean Theorem, it would be 2. So, factor out a 2: $\displaystyle 2\left(\dfrac{1}{2}\sin(x) + \dfrac{\sqrt{3}}{2}\cos(x)\right)$. Now, what angle has $\displaystyle \cos(c) = \dfrac{1}{2}$ and $\displaystyle \sin(c) = \dfrac{\sqrt{3}}{2}$? - Oct 25th 2013, 09:28 PMsepotoRe: Express in terms of Asin(x + c)
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My solutions manual describes I think a somewhat different solution. I'm thinking about the steps however I'm wondering what A and c are referring to. - Oct 26th 2013, 04:20 AMSlipEternalRe: Express in terms of Asin(x + c)
That is the exact same approach.

- Oct 26th 2013, 10:41 PMibduttRe: Express in terms of Asin(x + c)
In general let the expression be a sin P + b cos P. We can proceed by dividing and multiplying the expression by sqrt (a^2 + b^2) So we get

a sin P + b cos P = [sqrt (a^2 + b^2)] [ {a sin P + b cos P } / sqrt (a^2 + b^2) ] = [sqrt (a^2 + b^2)] [ a/{sqrt (a^2 + b^2)} sin P + b/{sqrt (a^2 + b^2)} cos P]

Now we can put a/sqrt (a^2 + b^2) = cos A then we will have b/sqrt (a^2 + b^2) = sin A and we will get

a sin P + b cos P = [sqrt (a^2 + b^2)] [ sin P cos A + sin A cos P ]= [sqrt (a^2 + b^2)] sin ( A + P)

Alternatively we can alternatively put

a/sqrt (a^2 + b^2) = sin A then we will have b/sqrt (a^2 + b^2) = cos A and we will get

a sin P + b cos P = [sqrt (a^2 + b^2)] [ sin P sin A + cos A cos P ]= [sqrt (a^2 + b^2)] cos ( P - A)