# Trigonometric Ratio question

• Jan 21st 2009, 03:45 PM
Random-Hero-
Trigonometric Ratio question
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.
• Jan 21st 2009, 03:48 PM
vincisonfire
Plug in the values
a)$\displaystyle 2sin(\frac{5\pi}{4})cos(\frac{3\pi}{4}) = 1$
b)$\displaystyle sin(\frac{8\pi}{4} + sin(\frac{2 \pi}{4}) = 1$
d) Do the same thing.
e) I guess that it is reasonable but it is not that hard to prove that it is true.
• Jan 21st 2009, 03:49 PM
o_O
This equation is true for all $\displaystyle x$ and $\displaystyle y$.

To prove it, start from the RHS and use the identities: $\displaystyle \sin (a \pm b) = \sin a \cos b \ \pm \ \cos a \sin b$
• Jan 21st 2009, 04:13 PM
Random-Hero-
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!
• Jan 21st 2009, 04:17 PM
o_O
Like I said, use the identities I gave you:

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

So, starting from the RHS:
\displaystyle \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!
• Jan 21st 2009, 04:31 PM
Random-Hero-
Wow thanks guys! That helped so much!! I really appreciate it!