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Rewriting a trigonometric expression of the form $A\\cos(\\theta)\\pm B\\sin(\\theta)$ to either $R\\sin(\\theta+\\alpha)$ or $R\\cos(\\theta+\\alpha)$ by calculating $R$ and $\\alpha$. 

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If

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{question}

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find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

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{answer}

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Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\var{R}. \\\\end{split} \\\\]

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\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\sqrt{\\\\var{A^2+B^2}}\\\\\\\\ &\\\\,=\\\\var{Rround} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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\\\\[ \\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} = \\\\simplify[unitFactor]{R sin (theta+{sign}*alpha)},\\\\]

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\\\\[ \\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} = \\\\simplify[unitFactor]{R cos (theta+{sign}*alpha)},\\\\]

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To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify[unitFactor]{sin(a+{sign}*b)=sin(a)cos(b)+{sign}*sin(b)cos(a)}$:

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\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} &\\\\,= \\\\simplify{R sin(theta+{sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(sin(theta)cos(alpha) + {sign}*sin(alpha)cos(theta))} \\\\\\\\ &\\\\,= \\\\simplify{Rsin(theta)cos(alpha) + {sign}*R sin(alpha)cos(theta)}. \\\\end{split} \\\\]

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By comparing the coefficients of $\\\\sin(\\\\theta)$ and $\\\\cos(\\\\theta)$, we find that

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\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

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To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

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{Rsol}

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Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

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\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

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Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify{cos(a+{sign}*b)=cos(a)cos(b)-{sign}*sin(a)sin(b)}$:

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\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} &\\\\,= \\\\simplify[unitFactor]{R cos (theta + {sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(cos(theta)cos(alpha) - {sign}*sin(theta)sin(alpha))} \\\\\\\\ &\\\\,= \\\\simplify{Rcos(theta)cos(alpha) - {sign}*R sin(theta)sin(alpha)}. \\\\end{split} \\\\]

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By comparing the coefficients of $\\\\cos(\\\\theta)$ and $\\\\sin(\\\\theta)$, we find that

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\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

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To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

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{Rsol}

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Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

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\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

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Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

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$R=$[[0]]

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$\\alpha=$[[1]]

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(Give your answers to 2 decimal places where necessary.)

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