Partial Solutions

1. Using \( \sin{(A-B)}=\sin{(A)}\cos{(B)} - \cos{(A)}\sin{(B)} \) where \( A=3x \) and \( B=x \),\[ \sin{(3x)}\cos{(x)}-\cos{(3x)}\sin{(x)}=\sin{(3x-x)}=\sin{(2x)} \]
2. Since \( \cos{(A+B)}=\cos{(A)}\cos{(B)} - \sin{(A)}\sin{(B)} \) then \( \sin{(A)}\sin{(B)} - \cos{(A)}\cos{(B)} = -\cos{(A+B)} \). If \( A= \dfrac{\pi}{5} \) and \( B= \dfrac{\pi}{3} \) then\[ \sin{\left(\dfrac{\pi}{5}\right)}\sin{\left(\dfrac{\pi}{3}\right)} - \cos{\left(\dfrac{\pi}{5}\right)}\cos{\left(\dfrac{\pi}{3}\right)} =-\cos{\left(\dfrac{\pi}{5} + \dfrac{\pi}{3}\right)}=-\cos{\left(\dfrac{8 \pi}{15}\right)} \]

3. Using \( \tan{(A+B)} = \dfrac{\tan{(A)} + \tan{(B)}}{1 - \tan{(A)}\tan{(B)}} \) and the fact that \( \cot{(\theta)} = \dfrac{1}{\tan{(\theta)}} \) (cotangent is the reciprocal of tangent), we have \( \cot{(A+B)} = \dfrac{1 - \tan{(A)}\tan{(B)}}{\tan{(A)} + \tan{(B)}} \).

   If \( A= 80^\circ \) and \( B= 20^\circ \), then \( \dfrac{1 - \tan{(80^\circ)}\tan{(20^\circ)}}{\tan{(80^\circ)} + \tan{(20^\circ)}} = \cot{(80^\circ + 20^\circ)} = \cot{(100^\circ)} \).

1.  \[\begin{align*} & \sin{\left(\dfrac{\pi}{4}-x\right)}-\sin{\left(\dfrac{5\pi}{4}+x\right)}\\ &= \sin{\left(\dfrac{\pi}{4}\right)}\cos{(x)}-\cos{\left(\dfrac{\pi}{4}\right)}\sin{(x)} - \left( \sin{\left(\dfrac{5\pi}{4}\right)}\cos{(x)} + \cos{\left(\dfrac{5\pi}{4}\right)}\sin{(x)} \right) \\ &= \left(\dfrac{1}{\sqrt{2}}\right)\cos{(x)} - \left(\dfrac{1}{\sqrt{2}}\right)\sin{(x)} - \left(-\dfrac{1}{\sqrt{2}}\right)\cos{(x)} - \left(-\dfrac{1}{\sqrt{2}}\right)\sin{(x)} \\ &= \dfrac{1}{\sqrt{2}}\cos{(x)} - \dfrac{1}{\sqrt{2}}\sin{(x)} + \dfrac{1}{\sqrt{2}}\cos{(x)} + \dfrac{1}{\sqrt{2}}\sin{(x)} \\ &= \dfrac{2}{\sqrt{2}}\cos{(x)}\\ &= \sqrt{2}\cos{\left(x\right)} \end{align*}\]Therefore, \( \sin{\left(\dfrac{\pi}{4}-x\right)}-\sin{\left(\dfrac{5\pi}{4}+x\right)} = \sqrt{2}\cos{(x)} \)
2.  \[\begin{align*} & \cos{\left(\theta+\dfrac{5\pi}{6}\right)}+\sin{\left(\theta -\dfrac{4\pi}{3}\right)}\\ &= \cos{\left(\theta\right)}\cos{\left(\dfrac{5\pi}{6}\right)}-\sin{\left(\theta\right)}\sin{\left(\dfrac{5\pi}{6}\right)} + \sin{(\theta)}\cos{\left(\dfrac{4\pi}{3}\right)} - \cos{(\theta)}\sin{\left(\dfrac{4\pi}{3}\right)} \\ &= \cos{\left(\theta\right)}\left(-\dfrac{\sqrt{3}}{2}\right)-\sin{\left(\theta\right)}\left( \dfrac{1}{2}\right) + \sin{(\theta)}\left(- \dfrac{1}{2}\right) - \cos{(\theta)}\left(-\dfrac{\sqrt{3}}{2}\right) \\ &= -\dfrac{\sqrt{3}}{2} \cos{(\theta)} - \dfrac{1}{2}\sin{(\theta)} - \dfrac{1}{2}\sin{(\theta)} + \dfrac{\sqrt{3}}{2}\cos{(\theta)} \\ &= -\sin{\left(\theta\right)} \end{align*}\]Therefore, \( \cos{\left(\theta+\dfrac{5\pi}{6}\right)}+\sin{\left(\theta -\dfrac{4\pi}{3}\right)} = -\sin{(\theta)} \).

From the diagrams below, \( \sin{(\beta)} = -\dfrac{\sqrt{8}}{3} \) and \( \cos{(\theta)} = -\dfrac{8}{17} \).

Now,

Therefore, \( \cos{(\theta+\beta)} = \dfrac{30~\sqrt{2}-8}{51} \).

1. \[\begin{align*} L.S. &= \cos{(\theta-\beta)}-\cos{(\theta+\beta)} \\ &= \cos{(\theta)}\cos{(\beta)} + \sin{(\theta)}\sin{(\beta)} - \left(\cos{(\theta)}\cos{(\beta)} - \sin{(\theta)}\sin{(\beta)}\right) \\ &= \cos{(\theta)}\cos{(\beta)} + \sin{(\theta)}\sin{(\beta)} - \cos{(\theta)}\cos{(\beta)} + \sin{(\theta)}\sin{(\beta)} \\ &= 2\sin{(\theta)}\sin{(\beta)} \\ R.S. &= 2\sin{(\theta)}\sin{(\beta)} \\ L.S. &= R.S. ~ \square \end{align*}\]
2. \[\begin{align*} L.S. &= \sin{(x+y)}\sin{(x-y)} \\ &= \left(\sin{(x)}\cos{(y)} + \cos{(x)}\sin{(y)}\right)\left(\sin{(x)}\cos{(y)} - \cos{(x)}\sin{(y)}\right) \\ &= \sin^2{(x)}\cos^2(y)-\cos^2(x)\sin^2(y) \\ &= \sin^2{(x)}[1-\sin^2{(y)}]-[1-\sin^2{(x)}]\sin^2{(y)} \\ &= \sin^2{(x)}-\sin^2{(x)}\sin^2{(y)}-\sin^2{(y)}+\sin^2{(x)}\sin^2{(y)}\\ &= \sin^2{(x)} - \sin^2{(y)} \\ R.S. &= \sin^2{(x)} - \sin^2{(y)} \\ L.S. &= R.S. ~ \square \end{align*}\]
3. \[\begin{align*} L.S. &= \cos{(x+y)}\sin{(x-y)} \\ &= \left(\cos{(x)}\cos{(y)} - \sin{(x)}\sin{(y)}\right)\left(\sin{(x)}\cos{(y)} - \cos{(x)}\sin{(y)}\right) \\ &= \sin{(x)}\cos{(x)}\cos^2{(y)}-\cos^2{(x)}\sin{(y)}\cos{(y)} - \sin^2{(x)}\sin{(y)}\cos{(y)} + \sin{(x)}\cos{(x)}\sin^2{(y)} \\ &= \sin{(x)}\cos{(x)}\cos^2{(y)} + \sin{(x)}\cos{(x)}\sin^2{(y)} - \cos^2{(x)}\sin{(y)}\cos{(y)} - \sin^2{(x)}\sin{(y)}\cos{(y)} \\ &= \sin{(x)}\cos{(x)}[\sin^2{(y)} + \cos^2{(y)}] - \sin{(y)}\cos{(y)}[\sin^2{(x)} + \cos^2{(x)}] \\ &= \sin{(x)}\cos{(x)} - \sin{(y)}\cos{(y)} \\ R.S. &= \sin{(x)}\cos{(x)} - \sin{(y)}\cos{(y)} \\ L.S. &= R.S. ~ \square \end{align*}\]
4. \[\begin{align*} L.S. &=\dfrac{\sin{(x+y)}}{\sin{(x-y)}}\\ &= \dfrac{\sin{(x)}\cos{(y)} + \cos{(x)}\sin{(y)}}{\sin{(x)}\cos{(y)} - \cos{(x)}\sin{(y)}} \\ &= \dfrac{\left(\sin{(x)}\cos{(y)} + \cos{(x)}\sin{(y)}\right)\div \cos{(x)}\cos{(y)}}{\left(\sin{(x)}\cos{(y)} - \cos{(x)}\sin{(y)}\right) \div \cos{(x)}\cos{(y)}} \\ &= \dfrac{\frac{\sin{(x)}\cancel{\cos{(y)}}}{\cos{(x)}\cancel{\cos{(y)}}} + \frac{\cancel{\cos{(x)}}\sin{(y)}}{\cancel{\cos{(x)}}\cos{(y)}}} {\frac{\sin{(x)}\cancel{\cos{(y)}}}{\cos{(x)}\cancel{\cos{(y)}}} - \frac{\cancel{\cos{(x}})\sin{(y)}}{\cancel{\cos{(x)}}\cos{(y)}}} \\ &= \dfrac{\tan{(x)}+\tan{(y)}}{\tan{(x)}-\tan{(y)}}\\ R.S. &= \dfrac{\tan{(x)}+\tan{(y)}}{\tan{(x)}-\tan{(y)}} \\ L.S. &= R.S. ~ \square \end{align*}\]
5. \[\begin{align*} L.S. =& \cot{(\theta)} - \cot{(\beta)} \\ =& \dfrac{\cos{(\theta)}}{\sin{(\theta)}} - \dfrac{\cos{(\beta)}}{\sin{(\beta)}} \\ =& \dfrac{\sin{(\beta)}\cos{(\theta)} - \cos{(\beta)}\sin{(\theta)}}{\sin{(\theta)}\sin{(\beta)}} \\ =& \dfrac{\sin{(\beta - \theta)}}{\sin{(\theta)}\sin{(\beta)}} \\ =& \dfrac{\sin{(-(\theta - \beta))}}{\sin{(\theta)}\sin{(\beta)}} \\ =& -\dfrac{\sin{(\theta - \beta)}}{\sin{(\theta)}\sin{(\beta)}} \\ R.S. =& -\dfrac{\sin{(\theta - \beta)}}{\sin{(\theta)}\sin{(\beta)}} \\ L.S. =& R.S. ~ \square \end{align*}\]
6. \[\begin{align*} L.S. &= \dfrac{\cot{(x)}\cot{(y)}-1}{\cot{(x)}+\cot{(y)}} \\ &= \dfrac{\dfrac{\cos{(x)}}{\sin{(x)}}\dfrac{\cos{(y)}}{\sin{(y)}}-1}{\dfrac{\cos{(x)}}{\sin{(x)}} + \dfrac{\cos{(y)}}{\sin{(y)}}} \\ &= \dfrac{\tfrac{\cos{(x)}\cos{(y)} - \sin{(x)}\sin{(y)}}{\cancel{\sin{(x)}\sin{(y)}}}}{\tfrac{\cos{(x)}\sin{(y)} + \sin{(x)\cos{(y)}}}{\cancel{\sin{(x)}\sin{(y)}}}} \\ &= \dfrac{\cos{(x)}\cos{(y)} - \sin{(x)}\sin{(y)}}{\cos{(x)}\sin{(y)} + \sin{(x)\cos{(y)}}} \\ &= \dfrac{\cos{(x+y)}}{\sin{(x+y)}} \\ &= \cot{(x+y)} \\ R.S. &= \cot{(x+y)} \\ L.S. &= R.S. ~ \square \end{align*}\]

1. First, express \( \sin{(x)} + \sqrt{3}\cos{(x)} \) in the form \( a\sin{(x-h)} \),

   \[\sin{(x)} + \sqrt{3}\cos{(x)} = a\sin{(x)}\cos{(h)} - a\cos{(x)}\sin{(h)}\]Equating coefficients,\[\begin{align*} a\cos{(h)} &= 1 \tag{1} \\ -a\sin{(h)} &= \sqrt{3} \tag{2} \end{align*}\] Dividing \( (2) \) by \( (1) \),\[\begin{align*} \dfrac{-a\sin{(h)}}{a\cos{(h)}} &= \dfrac{\sqrt{3}}{1} \\ -\tan{(h)} &= \dfrac{\sqrt{3}}{1} \\ \tan{(h)} &= -~\sqrt{3} \\ \end{align*}\] \( h = -\dfrac{\pi}{3} \) is one possible value; substituting this into \( (1) \),\[\begin{align*} a\cos{\left(-\dfrac{\pi}{3}\right)} &= 1 \\ a\left(\dfrac{1}{2}\right) &= 1 \\ a &= 2 \end{align*}\]

   \( \therefore \sin{(x)} + \sqrt{3}\cos{(x)} = 2\sin{\left(x+\dfrac{\pi}{3}\right)} \).

   Solving \( 2\sin{\left(x+\dfrac{\pi}{3}\right)} = \sqrt{2} \) for \( -\pi\leq x \leq 2\pi \), \( \sin{\left(x+\dfrac{\pi}{3}\right)} = \dfrac{\sqrt{2}}{2} \).

   Finding \( \left(x+\dfrac{\pi}{3}\right) \), where \( -\dfrac{2\pi}{3} \leq x+\dfrac{\pi}{3} \leq \dfrac{7\pi}{3} \), gives \( x+ \dfrac{\pi}{3} = \dfrac{\pi}{4}\), \(\dfrac{3\pi}{4}\), \(\dfrac{9\pi}{4} \). Therefore,

   \[\begin{align*} x &= \dfrac{\pi}{4} - \dfrac{\pi}{3}, \dfrac{3\pi}{4} - \dfrac{\pi}{3}, \dfrac{9\pi}{4} - \dfrac{\pi}{3} \\ x &= \dfrac{-\pi}{12}, \dfrac{5\pi}{12}, \dfrac{23\pi}{12} \\ &\approx -0.262, 1.309, 6.021 \end{align*}\]

   Verifying this solution using graphing technology,

2. First, express \( 4\sin{(x)} - 3\cos{(x)} \) in the form \( a\sin{(x-h)} \).

   \[ 4\sin{(x)} - 3\cos{(x)} = a\sin{(x)}\cos{(h)} - a\cos{(x)}\sin{(h)} \]

   Equating coefficients,

   \[\begin{align*} a\cos{(h)} &= 4 \tag{1} \\ a\sin{(h)} &= 3 \tag{2} \end{align*}\]

   Dividing \( (2) \) by \( (1) \),

   \[\begin{align*} \dfrac{a\sin{(h)}}{a\cos{(h)}} &= \dfrac{3}{4} \\ \tan{(h)} &= \dfrac{3}{4} \\ h &= \tan^{-1}{(0.75)} \\ \end{align*}\]

   \( h = 0.6435 \) is one possible value for \( h \); substituting this into \( (1) \),

   \[\begin{align*} a\cos{(0.6435)} &= 4 \\ a(0.8) &= 4 \\ a &= 5 \end{align*}\]

   \( \therefore 4\sin{(x)} - 3\cos{(x)}=5\sin{(x-0.6435)} \)

   Solving \( 5\sin{(x-0.6435)}=5 \) for \( 0\leq x \leq 2\pi \),

   \[\begin{align*} 5\sin{(x-0.6435)} &= 5 \\ \sin{(x-0.6435)} &= 1 \\ x- 0.6435 &= \dfrac{\pi}{2} \qquad (-0.6435 \leq x - 0.6435 \leq 5.6397) \\ x &= 2.214 \end{align*}\]

   Verifying this solution using graphing technology,

   

In \( \triangle AEF \), \( \angle EAF = \beta \) so \( \angle AFE = \dfrac{\pi}{2}-\beta \).

Since \( \sin(\beta) = \dfrac{EF}{1} \), then \( EF = \sin{(\beta)} \). Similarly, \( AE = \cos{(\beta)} \).

In \( \triangle ABE \), \( \angle BAE = \theta \) so \( \angle AEB = \dfrac{\pi}{2}-\theta \).

Since \( \sin{(\theta)} = \dfrac{BE}{AE} = \dfrac{BE}{\cos(\beta)} \), then \( BE = \sin{(\theta)}\cos{(\beta)} \). Similarly, \( AB = \cos{(\theta)}\cos{(\beta)} \).

In \( \triangle ADF \), \( \angle DAF = \dfrac{\pi}{2} - \theta - \beta \) and \( \angle AFD = \pi - \dfrac{\pi}{2}-\left(\dfrac{\pi}{2} - \theta - \beta \right) = \theta + \beta \).

Since \( \sin{(\theta + \beta)} = \dfrac{AD}{AF} = \dfrac{AD}{1} \), then \( AD = \sin{(\theta+\beta)} \). Similarly, we get \( DF = \cos{(\theta + \beta)} \).

In \( \triangle CEF \), \( \angle CEF = \pi - \dfrac{\pi}{2} - \left(\dfrac{\pi}{2}-\theta\right)= \theta \). Thus, \( \angle CFE = \dfrac{\pi}{2}-\theta \)

Since \( \sin{(\theta)} = \dfrac{CF}{EF} = \dfrac{BE}{\sin{(\beta)}} \), then \( CF = \sin{(\theta)}\sin{(\beta)} \). Similarly, \( CE = \cos{(\theta)}\sin{(\beta)} \).

In rectangle \( ABCD \), \( AD = BC = BE+CE \), \( \therefore \sin{(\theta + \beta)} = \sin{(\theta)}\cos{(\beta)} + \cos{(\theta)}\sin{(\beta)} \).

Also, \( AB = CD = CF + DF \), so

Therefore, \( \cos{(\theta + \beta)} = \cos{(\theta)}\cos{(\beta)} - \sin{(\theta)}\cos{(\beta)}\).