Trigonometric Ratio question

**Random-Hero-** · January 21st 2009, 03:45 PM

Consider the equation 2 Sin x cos y = sin ( x + y ) + sin ( x - y ).

Emilio claims that substituting x = 5pi/4 and y = 3pi/4 makes the equation true.

a) Determine an exact value for the left side using Emilio's solution.

b) Determine an exact value for the right side using Emilio's solution.

c) Is Emilio correct?

d) Is x = 5pi/6 and y = pi/6 a solution? Justify your answer.

e) Based on the answers to parts c) and d), is it reasonable to conclude that the equation is true no matter what angles are substituted for x and y? Justify your answer.

**vincisonfire** · January 21st 2009, 03:48 PM

Plug in the values
a) $2sin(\frac{5\pi}{4})cos(\frac{3\pi}{4}) = 1$
b) $sin(\frac{8\pi}{4} + sin(\frac{2 \pi}{4}) = 1$
d) Do the same thing.
e) I guess that it is reasonable but it is not that hard to prove that it is true.

**o_O** · January 21st 2009, 03:49 PM

This equation is true for all $x$ and $y$ .

To prove it, start from the RHS and use the identities: $\sin (a \pm b) = \sin a \cos b \ \pm \ \cos a \sin b$

**Random-Hero-** · January 21st 2009, 04:13 PM

I got most of the question, im just having a little trouble proving its true using the RHS LHS. Could someone help me break it down? It would help alot!

**o_O** · January 21st 2009, 04:17 PM

Like I said, use the identities I gave you:

${\color{red} \sin (a + b) = \sin a \cos b + \cos a \sin b}$
${\color{blue} \sin (a - b) = \sin a \cos b - \cos a \sin b}$

So, starting from the RHS:
$\begin{aligned} \text{RHS} & = {\color{red} \sin (x+ y)} + {\color{blue}\sin (x - y)} \\ & = {\color{red}\sin x \cos y + \cos x \sin y} + {\color{blue}\sin x \cos y - \cos x \sin y} \end{aligned}$

Now collect like terms!

**Random-Hero-** · January 21st 2009, 04:31 PM

Wow thanks guys! That helped so much!! I really appreciate it!

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