Hi my question is what approach i should use in order to solve this $\cos(\theta +20) = -0.74$?

and why do we get rid of the negative sign when solving basis angle?

Here is my working

$\cos(\theta +20) = -0.74$

Then, $\alpha+20 = \arccos (0.74)$, to find the basis angle. After I plugged into the calculator and got $42.3$, I subtract $20$ from $42.3$ to solve for $\theta$ which is $22.3$. $\theta = 42.3 -20 = 22.3$

Since $\cos$ is negative, the angle is at 2nd and 4th quadrant. So I solve for the angle $180-22.3=157.7$ and $180+22.3=202.3$

But all of this is wrong can someone explain? I think it is my approach that got me wrong, but I am unsure.

  • $\begingroup$ $20$ is probably not degrees. $\endgroup$ – Rebellos Nov 9 '18 at 13:05
  • $\begingroup$ Also remember that the equation actually has an infinite number of solutions. $\endgroup$ – Matti P. Nov 9 '18 at 13:06

$$\cos(\theta+20) = -0.74$$

$$\theta+20 = \arccos (-0.74)$$

Recall the range of $\arccos x$ is $0 \leq x \leq 180$.

$$\begin{cases} Quadrant \space II \implies \theta+20 = \arccos(-0.74) \implies \theta = \arccos(-0.74)-20\\ \ \\ Quadrant \space III \implies \theta+20 = 360-\arccos(-0.74) \implies \theta = 360-\arccos(-0.74)-20 \end{cases}$$

Finally, solve for $\theta$.

For the angle in the second quadrant:

$$\theta = \arccos(-0.74)-20 \approx 137.73-20 \approx 117.73$$

For the angle in the third quadrant:

$$\theta = 360-\arccos(-0.74)-20 \approx 360-137.73-20 \approx 202.27$$

Remember that angles repeat infinitely, so each angle repeats after $360°$. If the question asks for angles in the domain $0 \leq \theta \leq 360$, your answer is complete.

If, however, you’re asked for a complete set of solutions, you have to add in $\color{blue} {360n}$ for all $\color{blue}{n \in \mathbb{Z}}$.

$$\theta \approx 117.73+360n$$

$$\theta \approx 202.27+360n$$

By changing the order and signs, you reached an incorrect answer. Try to avoid doing that and you’ll be fine with these types of questions.

  • $\begingroup$ Thank you. I am actually preferred not to change the signs ie the negative sign and alpha symbol, but my teacher wanted me to use alpha to represent the basis angle which required to take away the negative sign $\endgroup$ – Fred Weasley Nov 9 '18 at 23:36
  • $\begingroup$ Good, I get your point. The reason you got the wrong answer using that method was that you calculated for $\theta$ first. You’re supposed to calculate the total angle, and then find $\theta$. Using your way, we can do the following: $$\arccos(0.74) \approx 42.27$$ Our total angle ($\theta+20$) is about $42.27°$. The total angle is in quadrants $2$ and $3$. Thus, for the second quadrant: $$\theta+20 \approx 180-42.27 \implies \theta+20 \approx 137.73 \implies \theta \approx 117.73$$ For the third: $$\theta+20 \approx 180+42.27 \implies \theta+20 \approx 222.27 \implies \theta \approx 202.27$$ $\endgroup$ – KM101 Nov 10 '18 at 12:02
  • $\begingroup$ So basically, you find the values for $\theta+20$, or rather, the entire angle, first. Then, you find $\theta$ for both situations. You used the reverse process, which got the signs mixed up. $\endgroup$ – KM101 Nov 10 '18 at 12:03
  • $\begingroup$ awesome thanks for helping out! But I have 1 more question, using the current way you have showed me, i tried in one of the questions and i have obtained a negative answer. What does this mean and how do i proceed? $tan(\theta-50)=-1.7$, solve for $\theta$ and i have got $-9.5$. Can you please help? Thank you and sorry $\endgroup$ – Fred Weasley Nov 10 '18 at 12:54
  • $\begingroup$ @Tfue $$\tan(\theta-50) = -1.7$$ $$\theta-50 = \arctan(-1.7)$$ Recall the range of $\arctan x$ is $-\frac{\pi}{2} < x < \frac{\pi}{2}$. So, the angle given will will be in the the fourth quadrant. But, tangent is also negative in the second quadrant. For the first case, $$\theta-50 = \arctan(-1.7) \implies \theta = \arctan(-1.7)+50 \implies \theta \approx -9.53446 \approx 350.46554$$ For the second case, $$\theta-50 = 180+\arctan(-1.7) \implies \theta = 180+\arctan(-1.7)+50 \implies \theta \approx 170.46554$$ $\endgroup$ – KM101 Nov 11 '18 at 8:55

In general we have that in degrees for $x\in[0,360°]$

$$\cos x = y \implies x=\arccos y \quad \lor \quad x=360°-\arccos y$$


$$\cos(\theta +20°) = -0.74 \implies \theta +20°=\arccos (-0.74) \quad \lor \quad \theta +20°=2\pi-\arccos (-0.74)$$

Check that the calculator is set to give the result in degree.

  • $\begingroup$ Thanks and what do i do when i got a negative angle? $\endgroup$ – Fred Weasley Nov 10 '18 at 1:26
  • $\begingroup$ Working in degrees we obtain $\arccos(-0.74)= 137.7$ then $\theta=157.7$ and $\theta=360-157.7=202.3$ are both solutions (when we set $\theta\in[0,360]$) and more in general any $\theta=\pm 157.7+360k$ with $k\in \mathbb{Z}$ is a possible solution. $\endgroup$ – user Nov 10 '18 at 2:29

Assuming all angles are in degrees, you should avoid strange switches in the sign.

With a calculator, $\arccos(-0.74)=137.7$, so the solutions are $$ \theta+20=137.7+360k $$ or $$ \theta+20=-137.7+360k $$ The solutions in the range $[0,360]$ are therefore $137.7-20=117.7$ or $-137-20+360=202.3$.

On the other hand, if angles are measured in radians, $$ \theta+20=\arccos(-0.74)+2k\pi=2.404+2k\pi $$ or $$ \theta+20=-\arccos(-0.74)+2k\pi=-2.404+2k\pi $$ Thus $$ \theta=-20+\arccos(-0.74)+2k\pi $$ or $$ \theta=-20-\arccos(-0.74)+2k\pi $$ For the solutions in the range $[0,2\pi]$ you have to choose $k=6$ in the first case and $k=8$ in the second case, obtaining $$ 1.253\qquad\text{or}\qquad 2.729 $$


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