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Exercise 3.3 · Q1

Q.Find the values of

(i) sin⁡(480∘)\sin(480^\circ)
(ii) sin⁡(−1110∘)\sin(-1110^\circ)
(iii) cos⁡(300∘)\cos(300^\circ)
(iv) tan⁡(1050∘)\tan(1050^\circ)
(v) cot⁡(660∘)\cot(660^\circ)
(vi) tan⁡(19π3)\tan\left(\dfrac{19\pi}{3}\right)
(vii) sin⁡(−11π3)\sin\left(-\dfrac{11\pi}{3}\right)
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✓ Free question

Every angle here reduces, after stripping off whole rotations of 360∘360^\circ (2π2\pi), to an allied angle of a standard acute angle; the table of §3.4.3 (or the negative-angle identity for a negative angle) then gives the value directly.

Step 1. Part (i): reduce 480∘480^\circ. 480∘=360∘+120∘480^\circ=360^\circ+120^\circ, and adding a full rotation never changes a trigonometric value, so sin⁡480∘=sin⁡120∘=sin⁡(180∘−60∘)=sin⁡60∘=32\sin480^\circ=\sin120^\circ=\sin(180^\circ-60^\circ)=\sin60^\circ=\dfrac{\sqrt3}{2}.

Step 2. Part (ii): reduce −1110∘-1110^\circ. First use sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta: sin⁡(−1110∘)=−sin⁡1110∘\sin(-1110^\circ)=-\sin1110^\circ. Since 1110∘=3×360∘+30∘1110^\circ=3\times360^\circ+30^\circ, sin⁡1110∘=sin⁡30∘=12\sin1110^\circ=\sin30^\circ=\dfrac12. So sin⁡(−1110∘)=−12\sin(-1110^\circ)=-\dfrac12.

Step 3. Part (iii): reduce 300∘300^\circ. 300∘=360∘−60∘300^\circ=360^\circ-60^\circ, so cos⁡300∘=cos⁡(−60∘)=cos⁡60∘=12\cos300^\circ=\cos(-60^\circ)=\cos60^\circ=\dfrac12 (using cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta, equally read off the table's 2π−θ2\pi-\theta column).

Step 4. Part (iv): reduce 1050∘1050^\circ. 1050∘=2×360∘+330∘1050^\circ=2\times360^\circ+330^\circ, so tan⁡1050∘=tan⁡330∘=tan⁡(360∘−30∘)=−tan⁡30∘=−13\tan1050^\circ=\tan330^\circ=\tan(360^\circ-30^\circ)=-\tan30^\circ=-\dfrac{1}{\sqrt3}.

Step 5. Part (v): reduce 660∘660^\circ. 660∘=360∘+300∘660^\circ=360^\circ+300^\circ, so cot⁡660∘=cot⁡300∘=cot⁡(360∘−60∘)=−cot⁡60∘=−13\cot660^\circ=\cot300^\circ=\cot(360^\circ-60^\circ)=-\cot60^\circ=-\dfrac{1}{\sqrt3}.

Step 6. Part (vi): reduce 19π3\dfrac{19\pi}{3}. 19π3=6π+π3\dfrac{19\pi}{3}=6\pi+\dfrac{\pi}{3}; stripping off 6π6\pi (three full rotations) leaves tan⁡19π3=tan⁡π3=3\tan\dfrac{19\pi}{3}=\tan\dfrac{\pi}{3}=\sqrt3.

Step 7. Part (vii): reduce −11π3-\dfrac{11\pi}{3}. sin⁡ ⁣(−11π3)=−sin⁡11π3\sin\!\left(-\dfrac{11\pi}{3}\right)=-\sin\dfrac{11\pi}{3}. Now 11π3=4π−π3\dfrac{11\pi}{3}=4\pi-\dfrac{\pi}{3}, so, dropping the 4π4\pi (two full rotations), sin⁡11π3=sin⁡ ⁣(−π3)=−sin⁡π3=−32\sin\dfrac{11\pi}{3}=\sin\!\left(-\dfrac{\pi}{3}\right)=-\sin\dfrac{\pi}{3}=-\dfrac{\sqrt3}{2}. Hence sin⁡ ⁣(−11π3)=−(−32)=32\sin\!\left(-\dfrac{11\pi}{3}\right)=-\left(-\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}.

✓Final answer

(i) 32\dfrac{\sqrt3}{2} (ii) −12-\dfrac12 (iii) 12\dfrac12 (iv) −13-\dfrac{1}{\sqrt3} (v) −13-\dfrac{1}{\sqrt3} (vi) 3\sqrt3 (vii) 32\dfrac{\sqrt3}{2}.

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