Can we solve cos θ − sin ( 2 θ ) + cos ( 3 θ ) − sin ( 4 θ ) = 0 \cos\theta - \sin (2\theta) + \cos (3\theta) - \sin (4\theta) = 0 cos θ − sin ( 2 θ ) + cos ( 3 θ ) − sin ( 4 θ ) = 0
Question
Give the general solutions of the following equations
2 sin 3 θ − 7 cos 2 θ + sin θ + 1 = 0 2\sin 3\theta - 7 \cos 2\theta + \sin \theta + 1 = 0 2 sin 3 θ − 7 cos 2 θ + sin θ + 1 = 0
cos θ − sin 2 θ + cos 3 θ − sin 4 θ = 0 \cos\theta - \sin 2\theta + \cos 3\theta - \sin 4\theta = 0 cos θ − sin 2 θ + cos 3 θ − sin 4 θ = 0
Suggestion
How might we reduce the number of different multiples of θ \theta θ
Can we turn the LHS into a function of sin θ \sin \theta sin θ cos θ \cos \theta cos θ
How can we simplify an expression like cos A + cos B \cos A + \cos B cos A + cos B
Solution
We’ll try to write everything in terms of just sin θ \sin \theta sin θ sin θ \sin\theta sin θ s = sin θ s = \sin \theta s = sin θ c = cos θ c = \cos\theta c = cos θ
We could use the double-angle formula cos 2 θ = 1 − 2 s 2 . \cos 2\theta = 1 - 2s^2. cos 2 θ = 1 − 2 s 2 . sin 3 θ \sin 3\theta sin 3 θ
sin 3 θ = c sin 2 θ + s cos 2 θ = 2 s c 2 + s ( 1 − 2 s 2 ) = 2 s ( 1 − s 2 ) + s ( 1 − 2 s 2 ) = 3 s − 4 s 3 . \begin{align*} \sin 3\theta &= c\sin 2\theta + s\cos 2\theta\\ &= 2sc^2 +s(1-2s^2)\\ &= 2s(1-s^2) + s(1-2s^2)\\ &= 3s - 4s^3. \end{align*} sin 3 θ = c sin 2 θ + s cos 2 θ = 2 s c 2 + s ( 1 − 2 s 2 ) = 2 s ( 1 − s 2 ) + s ( 1 − 2 s 2 ) = 3 s − 4 s 3 . The original equation becomes
2 ( 3 s − 4 s 3 ) − 7 ( 1 − 2 s 2 ) + s + 1 = 0 ⟹ 8 s 3 − 14 s 2 − 7 s + 6 = 0. \begin{align*} 2(3s - 4s^3) - 7(1-2s^2) + s +1 &= 0\\ \implies 8s^3 - 14s^2 - 7s + 6 &= 0. \end{align*} 2 ( 3 s − 4 s 3 ) − 7 ( 1 − 2 s 2 ) + s + 1 ⟹ 8 s 3 − 14 s 2 − 7 s + 6 = 0 = 0. When solving a cubic equation, it is often sensible to try simple numbers like s = ± 1 , ± 2 s= \pm 1, \pm 2 s = ± 1 , ± 2
If we try small integers for s s s 8 s 3 − 14 s 2 − 7 s + 6 = 0 8s^3 - 14s^2 - 7s + 6 = 0 8 s 3 − 14 s 2 − 7 s + 6 = 0 s = 2 s = 2 s = 2 s − 2 s-2 s − 2
8 s 3 − 14 s 2 − 7 s + 6 = ( s − 2 ) ( 8 s 2 + 2 s − 3 ) = ( s − 2 ) ( 4 s + 3 ) ( 2 s − 1 ) , 8s^3 - 14s^2 - 7s + 6 = (s-2)(8s^2+2s -3) =(s-2)(4s+3)(2s-1), 8 s 3 − 14 s 2 − 7 s + 6 = ( s − 2 ) ( 8 s 2 + 2 s − 3 ) = ( s − 2 ) ( 4 s + 3 ) ( 2 s − 1 ) , which is zero when s = 2 , − 3 4 s = 2, -\dfrac{3}{4} s = 2 , − 4 3 1 2 \dfrac{1}{2} 2 1
We can note that sin θ = 2 \sin \theta = 2 sin θ = 2
As the period of
sin \sin sin
is
2 π 2\pi 2 π
, and
sin ( θ ) = sin ( π − θ ) \sin(\theta) = \sin(\pi-\theta) sin ( θ ) = sin ( π − θ )
, the remaining solutions are
θ = sin − 1 ( 1 2 ) + 2 π n , θ = π − sin − 1 ( 1 2 ) + 2 π n , θ = sin − 1 ( − 3 4 ) + 2 π n , θ = π − sin − 1 ( − 3 4 ) + 2 π n \begin{align*} \theta = \quad & \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta = \quad & \pi - \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta =\quad & \sin^{-1}\left(- \frac 3 4\right) + 2\pi n, \\ \theta =\quad & \pi-\sin^{-1}\left(- \frac 3 4\right) + 2\pi n \end{align*} θ = θ = θ = θ = sin − 1 ( 2 1 ) + 2 π n , π − sin − 1 ( 2 1 ) + 2 π n , sin − 1 ( − 4 3 ) + 2 π n , π − sin − 1 ( − 4 3 ) + 2 π n where
n n n
is an integer. We can simplify to find
θ = π 6 + 2 n π , θ = 5 π 6 + 2 n π , θ = − sin − 1 ( 3 4 ) + 2 n π , θ = sin − 1 ( 3 4 ) + ( 2 n + 1 ) π , \begin{align*} \theta =\quad & \dfrac{\pi}{6} +2n\pi, \\ \theta =\quad & \dfrac{5\pi}{6} +2n\pi, \\ \theta =\quad & -\sin^{-1}\left(\frac 3 4\right)+2n\pi, \\ \theta =\quad & \sin^{-1}\left(\frac 3 4\right)+(2n+1)\pi, \end{align*} θ = θ = θ = θ = 6 π + 2 nπ , 6 5 π + 2 nπ , − sin − 1 ( 4 3 ) + 2 nπ , sin − 1 ( 4 3 ) + ( 2 n + 1 ) π , where in each case n n n
We might try an approach similar to that above, by writing everything in terms of cos θ \cos \theta cos θ sin θ \sin \theta sin θ sin \sin sin cos \cos cos
sin A + sin B = 2 sin ( A + B 2 ) cos ( A − B 2 ) \sin A + \sin B = 2 \sin\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right) sin A + sin B = 2 sin ( 2 A + B ) cos ( 2 A − B ) cos A + cos B = 2 cos ( A + B 2 ) cos ( A − B 2 ) . \cos A + \cos B = 2 \cos\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right). cos A + cos B = 2 cos ( 2 A + B ) cos ( 2 A − B ) .
We can derive these simply from
sin ( X ± Y ) = sin ( X ) cos ( Y ) ± cos ( X ) sin ( Y ) , cos ( X ± Y ) = cos ( X ) cos ( Y ) ∓ sin ( X ) sin ( Y ) \begin{align*} \sin(X\pm Y) &= \sin(X)\cos(Y) \pm \cos(X)\sin(Y),\\ \cos(X\pm Y) &= \cos(X)\cos(Y) \mp \sin(X)\sin(Y) \end{align*} sin ( X ± Y ) cos ( X ± Y ) = sin ( X ) cos ( Y ) ± cos ( X ) sin ( Y ) , = cos ( X ) cos ( Y ) ∓ sin ( X ) sin ( Y ) Applying these to this part of the question we obtain
cos θ + cos 3 θ = 2 cos 2 θ cos θ , sin 2 θ + sin 4 θ = 2 sin 3 θ cos θ \cos \theta + \cos 3\theta = 2 \cos 2\theta \cos \theta , \qquad \sin 2\theta + \sin 4\theta = 2 \sin 3\theta \cos \theta cos θ + cos 3 θ = 2 cos 2 θ cos θ , sin 2 θ + sin 4 θ = 2 sin 3 θ cos θ and so we need to solve
2 cos 2 θ cos θ − 2 sin 3 θ cos θ = 0 ⟺ cos θ ( cos 2 θ − sin 3 θ ) = 0. \begin{align*} 2\cos 2\theta \cos \theta - 2\sin3\theta\cos\theta & = 0 \\ \Longleftrightarrow \cos \theta (\cos 2 \theta - \sin 3\theta) & = 0. \end{align*} 2 cos 2 θ cos θ − 2 sin 3 θ cos θ ⟺ cos θ ( cos 2 θ − sin 3 θ ) = 0 = 0. So our solutions are cos θ = 0 \cos \theta = 0 cos θ = 0 θ = π 2 + n π \theta = \dfrac{\pi}{2} + n\pi θ = 2 π + nπ cos 2 θ = sin 3 θ \cos 2\theta = \sin 3\theta cos 2 θ = sin 3 θ
Now
cos 2 θ = 1 − 2 s 2 \cos 2\theta = 1-2s^2 cos 2 θ = 1 − 2 s 2
, while (as we showed above)
sin 3 θ = 3 s − 4 s 3 \sin 3\theta = 3s-4s^3 sin 3 θ = 3 s − 4 s 3
, so we need to solve
1 − 2 s 2 = 3 s − 4 s 3 ⟹ 4 s 3 − 2 s 2 − 3 s + 1 = 0 ⟹ ( s − 1 ) ( 4 s 2 + 2 s − 1 ) = 0 ⟹ sin θ = 1 or sin θ = − 1 ± 5 4 . \begin{align*} 1-2s^2&=3s-4s^3\\ \implies 4s^3-2s^2-3s+1&=0\\ \implies (s-1)(4s^2+2s -1)&=0\\ \implies \sin \theta = 1 \quad \text{or} \quad \sin \theta &= \dfrac{-1\pm\sqrt{5}}{4}. \end{align*} 1 − 2 s 2 ⟹ 4 s 3 − 2 s 2 − 3 s + 1 ⟹ ( s − 1 ) ( 4 s 2 + 2 s − 1 ) ⟹ sin θ = 1 or sin θ = 3 s − 4 s 3 = 0 = 0 = 4 − 1 ± 5 . So our final solution is
θ = π 2 + n π or θ = sin − 1 ( − 1 + 5 4 ) + 2 n π , or θ = π − sin − 1 ( − 1 + 5 4 ) + 2 n π , or θ = sin − 1 ( − 1 − 5 4 ) + 2 n π , or θ = π − sin − 1 ( − 1 − 5 4 ) + 2 n π , \begin{align*} \theta = \quad &\dfrac{\pi}{2} + n\pi\\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi, \\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi, \end{align*} θ = or θ = or θ = or θ = or θ = 2 π + nπ sin − 1 ( 4 − 1 + 5 ) + 2 nπ , π − sin − 1 ( 4 − 1 + 5 ) + 2 nπ , sin − 1 ( 4 − 1 − 5 ) + 2 nπ , π − sin − 1 ( 4 − 1 − 5 ) + 2 nπ , where n n n
Alternatively, we could note that cos α = cos β ⟹ α = ± β + 2 n π . \cos \alpha = \cos \beta \implies \alpha = \pm \beta + 2n\pi. cos α = cos β ⟹ α = ± β + 2 nπ .
We also have (by considering a right-angled triangle where one angle is γ \gamma γ sin γ = cos ( π 2 − γ ) . \sin \gamma = \cos\left(\dfrac{\pi}{2}-\gamma\right). sin γ = cos ( 2 π − γ ) . γ \gamma γ
Thus the equation
cos 2 θ = sin 3 θ \cos 2\theta = \sin 3\theta cos 2 θ = sin 3 θ
becomes
cos 2 θ = cos ( π 2 − 3 θ ) ⟹ 2 θ = ± ( π 2 − 3 θ ) + 2 n π ⟹ θ = π 2 + 2 n π or π 10 + 2 n π 5 , \begin{align*} \cos 2\theta &= \cos \left(\dfrac{\pi}{2}-3\theta \right)\\ \implies 2\theta &= \pm \left(\dfrac{\pi}{2}-3\theta\right) +2n\pi\\ \implies \theta &= \dfrac{\pi}{2} +2n\pi \quad \text{or} \quad \dfrac{\pi}{10} + \dfrac{2n\pi}{5}, \end{align*} cos 2 θ ⟹ 2 θ ⟹ θ = cos ( 2 π − 3 θ ) = ± ( 2 π − 3 θ ) + 2 nπ = 2 π + 2 nπ or 10 π + 5 2 nπ , and so we have the same answers as before, but this time in an explicit form.
Notice we have inadvertently shown here that sin ( π 10 ) = − 1 + 5 4 \sin\left(\dfrac{\pi}{10}\right) = \dfrac{-1+ \sqrt{5}}{4} sin ( 10 π ) = 4 − 1 + 5