UnkleRhaukus
  • UnkleRhaukus
\[f(x)=3\sin(x)-2\cos(x)\]
Trigonometry
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} f(x)&=3\sin(x)-2\cos(x)\\ &=\sqrt{(-2)^2+3^2}\cos\Big(x-\text{arctan2}\big({3,-2}\big)\Big)\\ &=\sqrt{13}\cos\Big(x-\big(\pi+\arctan(\tfrac3{-2})\big)\Big)\\ &=\sqrt{13}\cos\left(x-\pi+\arctan(\tfrac3{2})\right)\\ &=\sqrt{13}\sin\big(x-\tfrac\pi2+\arctan(\tfrac3{2})\big)\\ &\approx\sqrt{13}\sin\big(x-0.588\big)\\ \\ &A=\sqrt{13}\\ &\phi=-\frac\pi2+\arctan(\tfrac3{2})\approx -0.588\\ &T=2\pi\qquad\qquad\omega=\tfrac1{2\pi}\\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
right?

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ZeHanz
  • ZeHanz
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)
UnkleRhaukus
  • UnkleRhaukus
i can't see the typo
ZeHanz
  • ZeHanz
You wrote\[\arctan \left( \frac{ 3 }{ 2 } \right)\]It must be:\[\arctan \left( \frac{ 2 }{ 3 } \right)\]
UnkleRhaukus
  • UnkleRhaukus
i have applied these definitions, what method are you using
ZeHanz
  • ZeHanz
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....
UnkleRhaukus
  • UnkleRhaukus
i dont think i have made a mistake
ZeHanz
  • ZeHanz
If 3/2 was the same as 2/3 you didn't ;)
UnkleRhaukus
  • UnkleRhaukus
\[-\tfrac\pi2+\arctan(\tfrac32)\approx -0.588\] \[\qquad\quad\arctan(\tfrac23)\approx 0.588\]

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