# Solving $\tan\theta=\cos\theta$, for $-\pi <\theta < \pi$

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!

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

$$\tan\theta = \cos\theta \implies \sin\theta = \cos^2\theta = 1- \sin^2\theta \implies \sin^2\theta + \sin\theta -1 = 0\implies\sin\theta = \frac{-1\pm\sqrt{5}}{2}$$

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,

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

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

$$\tan t=\cos t\iff \cos^2t=\sin t\iff1-\sin^2t=\sin t\iff$$

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

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.

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 $$\sin\theta$$.

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