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Complete the table for the relation y = 3 x + 2 x for x = 30^, 60^, 90^, 120^, 150^, 210^. Round to 1 d.p.

Mathematics

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Step-by-step solution

8 steps

To complete the table for the relation y=3sin⁡x+2cos⁡xy = 3\sin x + 2\cos x, we will substitute each value of xx into the equation and calculate the corresponding yy value, rounding to 1 decimal place.

  1. Calculate yy for x=30∘x = 30^\circ.

    y=3sin⁡(30∘)+2cos⁡(30∘)=3(0.5)+2(32)=1.5+3≈1.5+1.732=3.232y = 3\sin(30^\circ) + 2\cos(30^\circ) = 3(0.5) + 2\left(\frac{\sqrt{3}}{2}\right) = 1.5 + \sqrt{3} \approx 1.5 + 1.732 = 3.232

    Rounded to 1 d.p., y=3.2y = 3.2.

  2. Calculate yy for x=60∘x = 60^\circ.

    y=3sin⁡(60∘)+2cos⁡(60∘)=3(32)+2(0.5)=332+1≈2.598+1=3.598y = 3\sin(60^\circ) + 2\cos(60^\circ) = 3\left(\frac{\sqrt{3}}{2}\right) + 2(0.5) = \frac{3\sqrt{3}}{2} + 1 \approx 2.598 + 1 = 3.598

    Rounded to 1 d.p., y=3.6y = 3.6.

  3. Calculate yy for x=90∘x = 90^\circ.

    y=3sin⁡(90∘)+2cos⁡(90∘)=3(1)+2(0)=3+0=3y = 3\sin(90^\circ) + 2\cos(90^\circ) = 3(1) + 2(0) = 3 + 0 = 3

    Rounded to 1 d.p., y=3.0y = 3.0.

  4. Calculate yy for x=120∘x = 120^\circ.

    y=3sin⁡(120∘)+2cos⁡(120∘)=3(32)+2(−12)=332−1≈2.598−1=1.598y = 3\sin(120^\circ) + 2\cos(120^\circ) = 3\left(\frac{\sqrt{3}}{2}\right) + 2\left(-\frac{1}{2}\right) = \frac{3\sqrt{3}}{2} - 1 \approx 2.598 - 1 = 1.598

    Rounded to 1 d.p., y=1.6y = 1.6.

  5. Calculate yy for x=150∘x = 150^\circ.

    y=3sin⁡(150∘)+2cos⁡(150∘)=3(0.5)+2(−32)=1.5−3≈1.5−1.732=−0.232y = 3\sin(150^\circ) + 2\cos(150^\circ) = 3(0.5) + 2\left(-\frac{\sqrt{3}}{2}\right) = 1.5 - \sqrt{3} \approx 1.5 - 1.732 = -0.232

    Rounded to 1 d.p., y=−0.2y = -0.2.

  6. Calculate yy for x=210∘x = 210^\circ.

    y=3sin⁡(210∘)+2cos⁡(210∘)=3(−0.5)+2(−32)=−1.5−3≈−1.5−1.732=−3.232y = 3\sin(210^\circ) + 2\cos(210^\circ) = 3(-0.5) + 2\left(-\frac{\sqrt{3}}{2}\right) = -1.5 - \sqrt{3} \approx -1.5 - 1.732 = -3.232

    Rounded to 1 d.p., y=−3.2y = -3.2.

  7. Calculate yy for x=300∘x = 300^\circ.

    y=3sin⁡(300∘)+2cos⁡(300∘)=3(−32)+2(0.5)=−332+1≈−2.598+1=−1.598y = 3\sin(300^\circ) + 2\cos(300^\circ) = 3\left(-\frac{\sqrt{3}}{2}\right) + 2(0.5) = -\frac{3\sqrt{3}}{2} + 1 \approx -2.598 + 1 = -1.598

    Rounded to 1 d.p., y=−1.6y = -1.6.

  8. Calculate yy for x=330∘x = 330^\circ.

    y=3sin⁡(330∘)+2cos⁡(330∘)=3(−0.5)+2(32)=−1.5+3≈−1.5+1.732=0.232y = 3\sin(330^\circ) + 2\cos(330^\circ) = 3(-0.5) + 2\left(\frac{\sqrt{3}}{2}\right) = -1.5 + \sqrt{3} \approx -1.5 + 1.732 = 0.232

    Rounded to 1 d.p., y=0.2y = 0.2.

    The completed table is:

    x0306090120150180210240270300330360y2.03.23.63.01.6−0.2−2.0−3.2−3.6−3.0−1.60.22.0\boxed{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 30 & 60 & 90 & 120 & 150 & 180 & 210 & 240 & 270 & 300 & 330 & 360 \\ \hline y & 2.0 & 3.2 & 3.6 & 3.0 & 1.6 & -0.2 & -2.0 & -3.2 & -3.6 & -3.0 & -1.6 & 0.2 & 2.0 \\ \hline \end{array} }
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