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 ^^
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Because Cos53.13 = 3/5 and sin53.13=4/5 (approximately.)
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 ^^ $\displaystyle cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) = cos(\alpha + \beta)$
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 ... $\displaystyle 5\left[\cos{\theta} \cdot \frac{3}{5} - \sin{\theta} \cdot \frac{4}{5}\right]$
Ya the same. But I think there is a typo in the actual question.
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
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