3
$\begingroup$

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!

$\endgroup$
3
  • 1
    $\begingroup$ Is the book asking for the value of $\sin \theta$ ? $\endgroup$ – Akash Karnatak Jul 13 '19 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 '19 at 10:26
  • $\begingroup$ If $sin \theta =\dfrac{-1+\sqrt{5}}{2}$ then $\theta=0.666\approx 0.67$ $\endgroup$ – Akash Karnatak Jul 13 '19 at 10:29
2
$\begingroup$

$\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,

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

$\endgroup$
2
  • 2
    $\begingroup$ You could also just use $\sinθ=-\sin(-θ)=\sin(π-θ)$ to get the second angle. $\endgroup$ – Lutz Lehmann Jul 13 '19 at 10:37
  • 1
    $\begingroup$ thanks! turns out my calculator was in degrees mode instead of radians.. serious schoolboy error! $\endgroup$ – Jay M Jul 13 '19 at 10:54
3
$\begingroup$

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

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

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.

$\endgroup$
0
2
$\begingroup$

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 $\sin\theta$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.