# Is this true?

• Feb 8th 2011, 03:11 PM
Googl
Is this true?
Hi, I am revising on Trig at the moment.

Would you say;

R sin (x+36.9) = R cos (x-53.1)

I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.
• Feb 8th 2011, 03:17 PM
dwsmith
Quote:

Originally Posted by Googl
Hi, I am revising on Trig at the moment.

Would you say;

R sin (x+36.9) = R cos (x-53.1)

I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.

Huh?

What was the original problem?

$\displaystyle 3\cos(x)-4\sin(x)=0\text{?}$

If so, solve

$\displaystyle \displaystyle\tan(x)=\frac{3}{4}$
• Feb 8th 2011, 03:22 PM
Googl
Quote:

Originally Posted by dwsmith
Huh?

What was the original problem?

$\displaystyle 3\cos(x)-4\sin(x)=0\text{?}$

If so, solve

$\displaystyle \displaystyle\tan(x)=\frac{3}{4}$

Would you be able to draw a cos and sin graph with that? Try to convert it to a cos/sin equation first and see what equation you end up with.
• Feb 8th 2011, 03:24 PM
dwsmith
Quote:

Originally Posted by Googl
Would you be able to draw a cos and sin graph with that? Try to convert it to a cos/sin equation first and see what equation you end up with.

What was the exact question you were given?
• Feb 8th 2011, 03:30 PM
Googl
Try to write $\displaystyle y = 3 cos x - 4 sin x$ as;

$\displaystyle y = R cos(x-[alpha])$

and

$\displaystyle y = R sin(x+[alpha])$

[alpha] is the alpha symbol

That
• Feb 8th 2011, 07:08 PM
Soroban
Hello, Googl!

Quote:

$\displaystyle \text{Try to write }y \:=\:3\cos x - 4\sin x$

$\displaystyle \text{as: }\;\begin{Bmatrix}y \:=\: R\cos(x-\alpha) \\ \text{and} \\ y \:=\: R\sin(x+\alpha) \end{Bmatrix}$

We have: .$\displaystyle y \;=\;3\cos x - 4\sin x$

Multiply by $\displaystyle \frac{5}{5}\!:$

. . $\displaystyle y \;=\;\frac{5}{5}(3\cos x - 4\sin x) \quad\Rightarrow\quad y \;=\; 5\left(\frac{3}{5}\cos x - \frac{4}{5}\sin x\right)$ .[1]

Let $\displaystyle \,\alpha$ be an acute angle is a 3-4-5 right triangle.

Code:

             *           /|           / |       5 /  |4         /  |       /α  |       *-----*         3

And we have: .$\displaystyle \begin{Bmatrix}\cos\alpha \:=\:\frac{3}{5} \\ \\[-3mm] \sin\alpha \:=\:\frac{4}{5} \end{Bmatrix}$

Substitute into [1]:

. . $\displaystyle y \;=\;5\left(\cos\alpha \cos x - \sin\alpha\sin x)$

Therefore: .$\displaystyle y \;=\;5\cos(x + \alpha)$

. . where $\displaystyle \alpha \:=\:\sin^{\text{-}1}\left(\frac{4}{5}\right) \:\approx\:53.1^o$

• Feb 8th 2011, 11:32 PM
Googl
I forgot to give you the R values sorry. R is 5 in my original post and it's similar to what you have here.

Would you say; $\displaystyle 5 sin (x+36.9) = 5 cos (x-53.1)$

Try to change it into the form;

R sin (x + [alpha])

alpha is the alpha symbol.
• Feb 9th 2011, 04:08 AM
HallsofIvy
Quote:

Originally Posted by Googl
Hi, I am revising on Trig at the moment.

Would you say;

R sin (x+36.9) = R cos (x-53.1)

36.9+ 53.1= 90.0 so if x is in degrees, yes, that is true. If we let y= x+ 36.9, then x- 53.1= y- 36.9- 53.1= y- 90.

But cos(y- 90)= cos(90- y) (since cosine is an even function) and it is always true that sin(y)= cos(90- y).

Quote:

I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.
• Feb 9th 2011, 02:19 PM
Googl
Thanks.