How to find angle x when sin(2x) is negative in quadrant IV

Key takeaway: This is a classic trig setup: first pin down where $2x$ sits, then back out $x$ carefully. The main trap is forgetting that halving an angle changes the quadrant logic.

Given that $2x$ is in quadrant IV and $\sin 2x=-\frac45$:

a) Locate $2x$

• In quadrant IV, sine is negative, which matches the given value.
• The reference triangle for $2x$ has:
  • hypotenuse $5$
  • opposite side $-4$
  • adjacent side $\sqrt{5^2-4^2}=3$

So on a sketch, place the angle in quadrant IV with a sine ratio of $-4/5$.

b) Find $x$ and its quadrant

Since $2x$ is in quadrant IV, one valid angle measure for $2x$ is $2x=360^\circ-\arcsin\left(\frac45\right).$

Using the reference angle: $\arcsin\left(\frac45\right)\approx 53.13^\circ,$ so $2x\approx 360^\circ-53.13^\circ=306.87^\circ.$ Then $x\approx \frac{306.87^\circ}{2}=153.44^\circ.$ That puts $x$ in quadrant II.

c) Find an exact value for $\sin x$

Use the half-angle identity: $\sin x=\pm\sqrt{\frac{1-\cos 2x}{2}}.$ From the triangle for $2x$, $\cos 2x=\frac{3}{5}.$ Because $x$ is in quadrant II, $\sin x$ is positive: $\sin x=\sqrt{\frac{1-\frac35}{2}}=\sqrt{\frac{\frac25}{2}}=\sqrt{\frac15}=\frac{\sqrt5}{5}.$

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Pitfalls the pros know 👇 A common mistake is to take $x$ in quadrant IV just because $2x$ is in quadrant IV. Halving the angle does not preserve quadrant automatically. Also, don’t use the negative square root for $\sin x$ here, because quadrant II means sine must be positive.

What if the problem changes? If the same trig ratio were given but $2x$ were in quadrant III instead, then $\sin 2x$ would still be negative, but $\cos 2x$ would be negative too. That would change the half-angle result for $\sin x$ and likely place $x$ in quadrant II or III depending on the interval you choose. The sign of $\sin x$ must always match the quadrant of $x$, not the quadrant of $2x$.

Tags: reference angle, half-angle identity, quadrant signs