1. ## simplify

5 [cosθ cos 53.13 deg - sinθ sin 53.13 deg] sinθ
=3cosθ -4sinθ

Can someone tell me how it arrive as 3cosθ -4sinθ ? Thanks ^^

2. Because Cos53.13 = 3/5 and sin53.13=4/5 (approximately.)

3. Originally Posted by jasonlewiz
5 [cosθ cos 53.13 deg - sinθ sin 53.13 deg] sinθ
=3cosθ -4sinθ

Can someone tell me how it arrive as 3cosθ -4sinθ ? Thanks ^^
$\displaystyle cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) = cos(\alpha + \beta)$

4. Originally Posted by jasonlewiz
5 [cosθ cos 53.13 deg - sinθ sin 53.13 deg] sinθ
=3cosθ -4sinθ

Can someone tell me how it arrive as 3cosθ -4sinθ ? Thanks ^^
$\displaystyle \cos(53.13^\circ) \approx \frac{3}{5}$

$\displaystyle \sin(53.13^\circ) \approx \frac{4}{5}$

left side of the equation ...

$\displaystyle 5\left[\cos{\theta} \cdot \frac{3}{5} - \sin{\theta} \cdot \frac{4}{5}\right]$

5. Ya the same.
But I think there is a typo in the actual question.

6. Originally Posted by bandedkrait
Ya the same.
But I think there is a typo in the actual question.
5 [cosθ cos 53.13 deg - sinθ sin 53.13 deg] sinθ
=3cosθ -4sinθ
yes ... the extra sinθ