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Determine the general solution to: $3 \sin \theta \cdot \sin 22^{\circ}=3 \cos \theta \cdot \cos 22^{\circ}+1$
in Mathematics by Diamond (81,058 points) | 41 views

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Answer:

$3 \sin \theta \cdot \sin 22^{\circ}=3 \cos \theta \cdot \cos 22^{\circ}+1$

$3 \sin \theta \cdot \sin 22^{\circ}-3 \cos \theta \cdot \cos 22^{\circ}=1 $
$-3\left(\cos \left(\theta+22^{\circ}\right)\right)=1 $
$\cos \left(\theta+22^{\circ}\right)=-\frac{1}{3} $
$\theta+22^{\circ}=\pm 109,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z} \quad $
$\theta=87,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z}$ or $\theta=-131,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z}$

 

Explanation:

In determining the general solution to $3 \sin \theta \cdot \sin 22^{\circ}=3 \cos \theta \cdot \cos 22^{\circ}+1$, there is
not an obvious route to the solution. You need to first establish how they can simplify the given trigonometric equation to a form where the sine of angle $=$ a number or the cosine of an angle $=$ a number or tan of an angle
= a number.
Now this invokes higher order reasoning and intuitive manipulations. Firstly, you need to transpose the trigonometric term $3 \cos \theta \cdot \cos 22^{\circ}$ to the LHS to give the following resultant equation: $3 \sin \theta \cdot \sin 22^{\circ}-3 \cos \theta \cdot \cos 22^{\circ}=1$.

You need to make a significant connection between the compound angle formula, $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$, and $3 \sin \theta \cdot \sin 22^{\circ}-$
$3 \cos \theta . \cos 22 .$ In order to do this you need to first recognize that $3 \sin \theta \cdot \sin 22^{\circ}-3 \cos \theta . \cos 22$ can be expressed in the following form:
$3\left(-\sin \theta \cdot \sin 22^{\circ}+\cos \theta \cdot \cos 22\right)=-3\left(\cos \theta \cdot \cos 22-\sin \theta \cdot \sin 22^{\circ}\right) .$ On succeeding
with this manipulation, you are most likely to see that he could apply the compound angle formulae $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$ to reduce
$\cos \theta . \cos 22^{\circ}-\sin \theta \cdot \sin 22^{\circ}$ to the form $\left(\cos \left(\theta+22^{\circ}\right)\right.$ and consequently arrive at
$-3\left(\cos \left(\theta+22^{\circ}\right)\right)=1$, which is significant step in path to finding the required
general solution. Thereafter, the candidate writes $-3\left(\cos \left(\theta+22^{\circ}\right)\right)=1$ in the
form $\cos \left(\theta+22^{\circ}\right)=-\frac{1}{3}$, which is in the form where

the cosine of an angle = a number. Now from here, you need to know that since cosine of the angle, $\left(\theta+22^{\circ}\right)$, is negative they must determine the reference angle (which in this case is $70,5^{\circ}$ ), and hence use it solve for $\theta$ in both quadrants 2 and 3, which will then enable them to derive the general solution as shown below:


$\theta+22=109,5+\mathrm{k} .360$ or $\theta+22=109,5+\mathrm{k} .360$
$\therefore \theta=87,5^{\circ}+k .360^{\circ}, k \in Z$ or $\therefore \theta=228,5^{\circ}+k \cdot 360^{\circ}, k \in Z$
Alternatively:
$\theta+22^{\circ}=\pm 109,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z}$
$\theta=87,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z}$ or $\theta=-131,5^{\circ}+\mathrm{k} .360^{\circ} ; \mathrm{k} \in \mathrm{Z}$


As discussed above, we see that this question requires higher order reasoning connected with the use of compound angles to help to reduce the given trigonometric equation to the form $\cos \left(\theta+22^{\circ}\right)=-\frac{1}{3}$, and the subsequent solving of the trigonometric equation, underpinned by good degree of conceptual understanding and higher order reasoning involving multiple steps. 

by Diamond (81,058 points)

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