Harmonic Form

In many instances it is essential to find the solutions of equations of the form


\(a \sin⁡ θ+b \cos⁡ θ=c\) (*) or
\(a \cos⁡θ+b \sin ⁡θ=c\) (**)
The previous methods for solving a trig equation cannot be applied directly to these equations. Therefore, we need to find an alternate form (of a single trig. ratio) of expressing the equation. This form is derived as follows:
From the diagram at left we see that
\[\sin⁡(θ+α)={c+d\over r}\]
\[⟹r \sin⁡ (θ+α)=c+d\]
Obtaining expressions for \(c\) and \(d\)
\[\sin ⁡θ={c\over a}\]
\[⟹c=a \sin ⁡θ\]

\[\cos ⁡θ={d\over b}\]
\[⟹d=b \cos⁡ θ\]
\[ \sin⁡(θ+α)=c+d\]
\[⟹r \sin⁡(θ+α)=a \sin ⁡θ+b \cos ⁡θ\]
The compares favourably to the left hand side of (*)
Also, we can get a similar expression for \(r \sin⁡(θ-α)\)
\[r \sin⁡(θ-α)=c-d\]
\[⟹r \sin⁡(θ-α)=a \sin ⁡θ-b \cos ⁡θ\]
Combing these two expressions we get
\[a \sin⁡ θ \pm b \cos ⁡θ=r \sin⁡(θ \pm α)\]
Similarly,
\[c=d-e\]
\[\cos⁡(θ+α)={c\over r}={d-e\over r}\]
\[⟹r \cos⁡(θ+α)=d-e\]
Obtaining expressions for \(d\) and \(e\)
\[\cos ⁡θ={d\over a}\]
\[⟹d=a \cos⁡θ\]
\[\sin⁡θ={e\over b}\]
\[⟹e=b \sin⁡θ\]

Therefore,
\[\cos⁡(θ+α)=a \cos⁡θ-b \sin⁡θ\]
Also,
\[r \cos⁡(θ-α)=d+e\]
\[r\cos \theta=a \cos⁡θ+b \sin⁡θ\]
Combining the two equations we get
\[r \cos⁡(θ∓α)=a \cos⁡θ\pm b \sin⁡θ\]
Furthermore, from the diagrams above we determine that
\[r=\sqrt{a^2+b^2}\] and \[α=\tan^{-1} ({b\over a})\]
NB: The absolute values of \(a\) and \(b\) are to be used in the above calculations.
Summarising we have
\[a \sin⁡θ \pm b \cos⁡θ=\sqrt{a^2+b^2} \sin⁡(θ\pm α); α=\tan^{-1}⁡({b\over a})\]
\[a \cos⁡θ\pm b \sin⁡θ=\sqrt{a^2+b^2} \cos⁡(θ∓α); α=\tan^{-1}⁡({b\over a})\]
Thus, we are now equipped to solve the required equations.

LESSON 1
Solve the equation \(3 \sin⁡θ-4 \cos⁡θ=1\) for \(0°≤θ≤360°\)
SOLUTION
Choose whether you want to express the left hand side in the form \(R \sin⁡(θ±α)\) or \(r \cos⁡(θ∓α)\).
The equation is in the form a \(\sin⁡θ-b \cos⁡θ=1\).
Using \(r \sin⁡(θ-α)\) we have to note that \(a\) is the coefficient of \(\sin⁡θ\) and \(b\) is the coefficient of \(\cos⁡θ\)
Determine the value of \(r\)
\[r=\sqrt{3^2+4^2}=5\]
Determine the value of \(α\)
\[α=\tan^{-1}⁡{4\over3}=53.13°\]
Substitute values
\[3 \sin⁡θ-4 \cos⁡θ=r \sin⁡(θ-α)\]
\[⟹3 \sin⁡θ-4 \cos⁡θ=5 \sin⁡(θ-53.13°)\]
Rewrite original equation
\[3 \sin⁡θ-4 \cos⁡θ=1\]
\[⟹5 \sin⁡(θ-53.13°)=1\]
Let \(x=θ-53.13°\)
\[5 \sin⁡ x=1\]
Solve for \(x\)
\[\sin ⁡x={1\over5}\]
\[RA=\sin^{-1}({1\over5})=11.54°\]
Sine is positive in I and II
\[I⟹x=11.54°\]
\[II⟹x=180°-11.54°\]
\[⟹x=168.46°\]
Solve for \(θ\)
\[θ-53.13°=11.54°\]
\[⟹θ=11.54°+53.13°\]
\[⟹θ=64.67°\]
\[θ-53.13°=168.46°\]
\[⟹θ=168.46°+53.13°\]
\[⟹θ=221.59°\]

METHOD 2

Rewrite equation with the coefficient of \(\cos⁡θ\) as positive
\[-4 \cos⁡θ+3 \sin⁡θ=1\]
\[⟹4 \cos⁡θ-3 \sin⁡θ=-1\]
Alternately, using \(r \cos⁡(θ+α)\) we have to note that \(a\) is the coefficient of \(\cos⁡θ\) and \(b\) is the coefficient of \(\sin⁡θ\)
Determine the value of \(r\)
\[r=\sqrt{4^2+3^2}=5\]
Determine the value of \(α\)
\[α=\tan^{-1}({3\over4})=36.87°\]
Substitute values
\[4 \cos⁡θ-3 \sin⁡θ=r \cos⁡(θ+α)\]
\[⟹4 \cos⁡θ-3 \sin⁡θ=5 \cos⁡(θ+36.87°)\]
Rewrite original equation
\[4 \cos⁡θ-3 \sin⁡θ=-1\]
\[⟹5 \cos⁡(θ+36.87°)=-1\]
Let \(x=θ+36.87°\)
\[5 \cos ⁡x=-1\]
Solve for \(x\)
\[\cos⁡ x=-{1\over5}\]
\[RA=\cos^{-1}⁡({1\over5})=78.46°\]
Cosine is negative in II and III
\[II⟹x=180°-78.46°\]
\[⟹x=101.54°\]
\[III⟹x=180°+78.46°\]
\[⟹x=258.46°\]
Solve for \(θ\)
\[θ+36.87°=101.54°\]
\[⟹θ=101.54°-36.87°\]
\[⟹θ=64.67°\]

\[θ+36.87°=258.46°\]
\[⟹θ=258.46°-36.87°\]
\[⟹θ=221.59°\]

TRY THIS

1. Express \(\cos⁡θ+√3 \sin⁡θ\) in the form \(R \cos⁡(θ-α)\), where \(R>0\) and \(α\) is acute, expressing \(α\) in terms of \(π\).
Ans: \(2 \cos⁡(θ-{π\over3})\)
2. (i) Express \(5 \cos⁡ x+12 \sin ⁡x\) in the form \(R \cos⁡(x-α)\), where \(R>0\) and \(0°<α<90°\).
(ii) Solve, for \(0°<x<360°\), the equation \(5 \cos ⁡x+12 \sin ⁡x=2\), giving your answers correct to the nearest \(0.1°\).
Ans: (i) \(13 \cos⁡(x-67.4°)\) (ii) \(346.2°\)
3. (i) Express \(4 \cos⁡θ-\sin⁡θ\) in the form \(R \cos⁡(θ+α)\), where \(R>0\) and \(0°<α<90°\).
(ii) Hence solve the equation \(4 \cos⁡θ-\sin⁡θ=2\), giving all solutions for which \(-180°<θ<180°\).
Ans: (i) \(\sqrt{17} \cos⁡(θ+14°)\) (ii) \(-75°, 46.9°\)
4. (i) Express \(24 \sin⁡θ+7 \cos⁡θ\) in the form \(R \sin⁡(θ+α)\), where \(R>0\) and \(0°<α<90°\).
(ii) Hence solve the equation \(24 \sin⁡θ+7 \cos⁡θ=12\) for \(0°<θ<360°\).
Ans: (i) \(25 \sin⁡(θ+16.3°)\) (ii) \(12.4°,135.1°\)

5. Express \(\sin⁡θ-3 \cos⁡θ\) in the form \(R \sin⁡(θ-α)\), where \(R\) and \(α\) are constants to be determined, and \(0°<α<90°\).
Hence solve the equation \(\sin⁡θ-3 \cos⁡θ=1\) for \(0°\leq θ\leq 360°\).
Ans: \(\sqrt{10} \sin⁡(θ-71.57°)\), \(θ=90°,233.1°\)

LESSON 2
Express \(f(θ)=\sqrt{2} \cos⁡θ+\sin⁡θ\) in the form \(R \cos(θ-α)\) where \(R>0\) and \(α\) is acute.
Hence, find the minimum value of \(f(θ)\),where \(0\leq θ\leq 2π\).
Determine the value of \(θ, 0\leq θ\leq 2π\), at which the minimum value of \(f(θ)\) occurs.
SOLUTION
Determine the value of \(r\)
\[r=\sqrt{(\sqrt{2})^2+1^2}=\sqrt{3}\]
Determine the value of \(α\)
\[α=\tan^{-1}⁡({1\over \sqrt{2}})=0.615\]
Substitute values
\[\sqrt{2} \cos⁡θ+\sin⁡θ=\sqrt{3} \cos⁡(θ-0.615)\]
We know that the minimum value of the cosine function is \(-1\). Hence, the minimum value of
\[f(θ)=\sqrt{3} \cos⁡(θ-0.615)\]
\[f(θ)\sqrt{3} (-1)=-\sqrt{3}\]
In general, the maximum value of \(f(θ)=r \sin⁡(θ±α)\) and \(f(θ)=r \cos⁡(θ∓α)\) is \(r\) and the corresponding minimum value is \(–r\).
Since the minimum value occurs when \(\cos⁡(θ-0.615)=-1\) we need to solve the equation
\[\cos⁡(θ-0.615)=-1\]
Let \(x=θ-0.615\)
\[\cos⁡x=-1\]
From the cosine graph we determine that \(\cos⁡x=-1\) when \(x=π\). Alternately, the RA can be used.
\[x=π\]
\[⟹θ-0.615=π\]
\[⟹θ=π+0.615\]
\[⟹θ=3.76\]

LESSON 3
Express \(f(2θ)=4 \sin⁡2θ+3 \cos⁡2θ\) in the form \(r \sin⁡(2θ+α)\) where \(r>0\) and \(0<α<{π\over2}\).
Hence, or otherwise, find the maximum and minimum values of
\[{1\over6-f(θ)}\]
SOLUTION
\[4 \sin⁡2θ+3 \cos⁡2θ\]
\[r=\sqrt{4^2+3^2}=5\]
\[α=\tan^{-1}⁡({3\over4})=0.644^c\]
\[f(2θ)=5 \sin⁡(2θ+0.644)\]
The minimum and maximum values of \(f(2θ)\) and f\((θ)\) are the same.
Maximum value of \(f(θ)=5\)
Minimum value of \(f(θ)=-5\)

Maximum value of
\[{1\over 6-f(θ)}={1\over 6-5}=1\]
Minimum value of
\[{1\over 6-f(θ)}={1\over 6-(-5)}={1\over11}\]

TRY THIS

1. Express \(2 \sin⁡θ-3 \cos⁡θ\) in the form \(R \sin⁡(θ-α)\), where \(R\) and \(α\) are constants to be determined, and \(0<α<{π\over2}\).

Hence write down the greatest and least values of \(1+2 \sin⁡θ-3 \cos⁡θ\).
Ans: \(√13 \sin⁡(θ-0.983)\),min:\(1-√13\),max:\(1+√13\)
2. (i) Express \(3 \sin⁡θ+4 \cos⁡θ\) in the form \(R \sin⁡(θ+α)\), where \(R>0\) and \(0°<α<90°\).
(ii) Hence
(a) solve the equation \(3 \sin⁡θ+4 \cos⁡θ+1=0\), giving all solutions for which \(-180°<θ<180°\).
(b) find the values of the positive constants \(k\) and \(c\) such that
\[-37≤k(\sin⁡θ+4 \cos⁡θ )+c≤43\]
for all values of \(θ\).
Ans: (i) \(5 \sin⁡(θ+53.1°)\) (ii) (a) \(-64.7°,138°\) (b) \(k=8, c=3\)
3. (i) Express \(4 \cos⁡θ-2 \sin⁡θ\) in the form \(R \cos⁡(θ+α)\), where \(R>0\) and \(0°<α<90°\).
(ii) Hence
(a) solve the equation \(4 \cos⁡θ-2 \sin⁡θ=3\) for \(0°<θ<360°\).
(b) determine the greatest and least values of
\[25-(4 \cos⁡θ-2 \sin⁡θ )^2\]
as \(θ\) varies, and, in each case, find the smallest positive value of \(θ\) for which that value occurs.
Ans: (i) (a) \(√20 \cos⁡(θ+26.6°)\) (ii) (a) \(21.3°,285.6°\) (b) Max\(=1,θ=63.4°\) Min\(=5,θ=153.4°\)
4. Express \(3 \cos⁡θ+4 \sin⁡θ\) in the form \(R \cos⁡(θ-α)\), where \(R>0\) and \(0<α<{π\over2}\).
Hence find the range of the function \(f(θ)\), where \(f(\theta)=7+3 \cos⁡θ+4 \sin⁡θ\) for \(0≤θ≤2π\)
Write down the greatest possible value of
\[1\over 7+3 \cos⁡θ+4 \sin⁡θ\]
Ans: \(5 \cos⁡(θ-0.927)\) Range: 2 to 12 Greatest value is \(\displaystyle {1\over2}\)
5. Solve the equation \(2 \sin⁡2θ+\cos⁡2θ=1\), for \(0°\leq θ<360°\).
Ans: \(θ=63.43°,243.43°\)