Question

Exact values of trig expressions using reference angles and special angles

Original question: 1. Express each of the following as a function of its co-related acute angle and evaluate. a) cos 330° b) sin − 210° c) sin 120° d) tan (− 300°) e) tan 510° f) cos (− 315°)

Expert Verified Solution

thumb_up100%(1 rated)

Expert intro: These are classic special-angle questions. The main trick is to reduce each angle to a reference angle and keep the sign from the quadrant.

Detailed walkthrough

Use reference angles and quadrant signs.

a) cos330\cos 330^\circ

330=36030330^\circ=360^\circ-30^\circ, so

cos330=cos30=32.\cos 330^\circ=\cos 30^\circ=\frac{\sqrt{3}}{2}.

b) sin(210)\sin(-210^\circ)

Use odd symmetry:

sin(θ)=sinθ.\sin(-\theta)=-\sin\theta.

So

sin(210)=sin210.\sin(-210^\circ)=-\sin 210^\circ.

Now 210=180+30210^\circ=180^\circ+30^\circ, hence

sin210=sin30=12.\sin 210^\circ=-\sin 30^\circ=-\frac12.

Therefore,

sin(210)=12.\sin(-210^\circ)=\frac12.

c) sin120\sin 120^\circ

120=18060120^\circ=180^\circ-60^\circ, so

sin120=sin60=32.\sin 120^\circ=\sin 60^\circ=\frac{\sqrt{3}}{2}.

d) tan(300)\tan(-300^\circ)

tan\tan is odd, so

tan(300)=tan300.\tan(-300^\circ)=-\tan 300^\circ.

And 300=36060300^\circ=360^\circ-60^\circ, so

tan300=tan60=3.\tan 300^\circ=-\tan 60^\circ=-\sqrt{3}.

Thus,

tan(300)=3.\tan(-300^\circ)=\sqrt{3}.

e) tan510\tan 510^\circ

Subtract 360360^\circ:

510360=150,510^\circ-360^\circ=150^\circ,

so

tan510=tan150.\tan 510^\circ=\tan 150^\circ.

Now 150=18030150^\circ=180^\circ-30^\circ, hence

tan150=tan30=13=33.\tan 150^\circ=-\tan 30^\circ=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}.

f) cos(315)\cos(-315^\circ)

cos\cos is even, so

cos(315)=cos315.\cos(-315^\circ)=\cos 315^\circ.

And 315=36045315^\circ=360^\circ-45^\circ, therefore

cos315=cos45=22.\cos 315^\circ=\cos 45^\circ=\frac{\sqrt{2}}{2}.

Answers

  • a) 32\frac{\sqrt{3}}{2}
  • b) 12\frac12
  • c) 32\frac{\sqrt{3}}{2}
  • d) 3\sqrt{3}
  • e) 33-\frac{\sqrt{3}}{3}
  • f) 22\frac{\sqrt{2}}{2}

💡 Pitfall guide

The biggest trap is sign handling. Students often find the reference angle correctly but forget whether the angle is in quadrant II, III, or IV. Also, for negative angles, use symmetry first if that makes the work cleaner; it prevents accidental sign flips.

🔄 Real-world variant

If the same expressions were written in radians, the method would not change: reduce to a standard angle, use even/odd identities, then apply the known special-angle values. The exact values would still come from π/6\pi/6, π/4\pi/4, and π/3\pi/3 type angles.

🔍 Related terms

reference angle, unit circle, quadrant signs

FAQ

How do I find exact trig values from angles like 330° or 510°?

Reduce the angle to a related acute reference angle, identify the quadrant, and apply the correct sign using unit-circle rules.

Why is sin of a negative angle sometimes positive?

Because sine is an odd function. For example, sin(-θ) = -sin(θ). If sin(θ) is already negative, the double negative makes the result positive.

chat