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Solve for $-\pi <\theta < \pi$: $$\tan\theta=\cos\theta$$

I can't get to the correct solution using the identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta} \quad\text{and}\quad \sin^2\theta+\cos^2\theta=1$$

The answer I'm getting is

$$\sin\theta=-\frac12\pm\frac12 \sqrt{5}$$

giving: $0.62$ and $-1.62$.

The answers in the back of the book are $0.67$ and $2.48$.

Any hints much appreciated. Thanks!

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    $\begingroup$ Is the book asking for the value of $\sin \theta$ ? $\endgroup$ – Akash Karnatak Jul 13 at 10:24
  • $\begingroup$ 1) $\sin\theta$ cannot be negative, being equal to a square, and 2) $\sin\theta = (-1+\sqrt5)/2$ has two solutions for $\theta \in [0, \pi]$. $\endgroup$ – WimC Jul 13 at 10:26
  • $\begingroup$ If $sin \theta =\dfrac{-1+\sqrt{5}}{2}$ then $\theta=0.666\approx 0.67$ $\endgroup$ – Akash Karnatak Jul 13 at 10:29
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$\tan\theta = \cos\theta \implies \sin\theta = \cos^2\theta = 1- \sin^2\theta \implies \sin^2\theta + \sin\theta -1 = 0\implies\sin\theta = \frac{-1\pm\sqrt{5}}{2} $

But $-1\le\sin\theta \le1$

So,

$$\sin\theta = \frac{-1+\sqrt5}{2} \approx 0.618$$

$$\theta = \sin^{-1}0.618 = 0.667 \ rad, $$

Also $$\sin\theta = \cos(\pi/2-\theta) = \cos(\theta-\pi/2)= 0.618$$

$$\theta - 1.57 = 0.904 \implies \theta = 2.474 \ rad$$

I've not used $\cos(\pi/2-\theta) = 0.618$ as it'd lead to to the first solution.

As commented by @LutzL,

$$\sin(\pi-\theta) = \sin\theta = 0.618\implies \pi-\theta = 0.667 \implies \theta = \pi -0.667 \approx 2.274$$

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    $\begingroup$ You could also just use $\sinθ=-\sin(-θ)=\sin(π-θ)$ to get the second angle. $\endgroup$ – LutzL Jul 13 at 10:37
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    $\begingroup$ thanks! turns out my calculator was in degrees mode instead of radians.. serious schoolboy error! $\endgroup$ – Jay M Jul 13 at 10:54
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$$\tan t=\cos t\iff \cos^2t=\sin t\iff1-\sin^2t=\sin t\iff$$

$$\sin^2t+\sin t-1=0\iff \sin t=\frac{-1\pm\sqrt5}2$$

But $\;|\sin t|\le1\;$ always, so we're left with only

$$\sin t=\frac{-1+\sqrt5}2\implies t=\begin{cases}\arcsin\frac{-1+\sqrt5}2+2k\pi=0.667 Rad.+2k\pi\\{}\\\pi-\arcsin\frac{-1+\sqrt5}2+2k\pi\end{cases}\,,\,\,k\in\Bbb Z$$

You have one answer above. and you can now check yourself what the other one is.

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If $\cos\theta=\tan\theta=\tfrac{\sin\theta}{\cos\theta}$ then $$\sin\theta=\cos^2\theta=1-\sin^2\theta,$$ which gives a quadratic equation in $\sin\theta$. The quadratic formula then gives $$\sin\theta=-\frac12\pm\frac12\sqrt{5},$$ as you found. Note that the negative solution is impossible because we started off with $\sin\theta=\cos^2\theta$.

This shows that $\sin\theta=-\tfrac12+\tfrac12\sqrt{5}$, and correspondingly either $$\theta\approx0.666\qquad\text{ or }\qquad\theta\approx2.475,$$ so perhaps the question asked for the value of $\theta$, not $\sin\theta$.

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What you are asked is the value of $\color{blue}\theta$ and not $\color{blue}\sin \theta$

Using what you have already written, $\sin \theta = \cos ^2 \theta = 1 - \sin ^2 \theta$. Write $x = \sin \theta$, the equation is now $x^2 + x - 1 = 0$, you will get two values of $x$, one of them will be less than $-1$ which you have already mentioned and we can discard, the other is $\dfrac{\sqrt 5 - 1}{2} = 0.618$ which you had already obtained.

Now the only question is: For what angles is $\sin \theta = 0.618$. And look up the inverse sine table or use a calculator to get the answers you are looking for

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