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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\begin{align*} f(x)&=3\sin(x)2\cos(x)\\ &=\sqrt{(2)^2+3^2}\cos\Big(x\text{arctan2}\big({3,2}\big)\Big)\\ &=\sqrt{13}\cos\Big(x\big(\pi+\arctan(\tfrac3{2})\big)\Big)\\ &=\sqrt{13}\cos\left(x\pi+\arctan(\tfrac3{2})\right)\\ &=\sqrt{13}\sin\big(x\tfrac\pi2+\arctan(\tfrac3{2})\big)\\ &\approx\sqrt{13}\sin\big(x0.588\big)\\ \\ &A=\sqrt{13}\\ &\phi=\frac\pi2+\arctan(\tfrac3{2})\approx 0.588\\ &T=2\pi\qquad\qquad\omega=\tfrac1{2\pi}\\ \end{align*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
right?
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
It's\[f(x)=\sqrt{3^2+(2)^2}\sin \left( x\arctan \left( \frac{ 2 }{ 3 } \right) \right)\]\[\approx \sqrt{13}\sin (x 0.588)\]So you are right, but you made a typo in the equation editor (3/2 instead of 2/3)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i can't see the typo
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
You wrote\[\arctan \left( \frac{ 3 }{ 2 } \right)\]It must be:\[\arctan \left( \frac{ 2 }{ 3 } \right)\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i have applied these definitions, what method are you using
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
I applied the same rule. Your answer is right. You just made a typo in your explanation above (in the equation editor). Just carefully look what you typed there (3/2). In your actual calculation you used 2/3, which is right....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i dont think i have made a mistake
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
If 3/2 was the same as 2/3 you didn't ;)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\tfrac\pi2+\arctan(\tfrac32)\approx 0.588\] \[\qquad\quad\arctan(\tfrac23)\approx 0.588\]
 one year ago
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