## 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: $$θ=63.43°,243.43°$$