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Exercise 3.2 · Q6

Q.Find the value of the trigonometric function sin⁡765∘\sin 765^\circ.

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Since sin⁡\sin has period 360∘360^\circ, sin⁡765∘=sin⁡(765∘−2×360∘)=sin⁡45∘=12\sin 765^\circ = \sin(765^\circ - 2\times360^\circ) = \sin 45^\circ = \dfrac{1}{\sqrt2}.

Why this works: periodicity of sine

The sine function repeats every 360∘360^\circ — a full rotation returns to the same point on the unit circle. So for a large angle like 765∘765^\circ, first strip off as many full 360∘360^\circ turns as possible; the sine of what's left is the answer.

Step-by-step solution

Step 1 — Reduce the angle to within one revolution.

765∘−2×360∘=765∘−720∘=45∘765^\circ - 2\times360^\circ = 765^\circ - 720^\circ = 45^\circ

So sin⁡765∘=sin⁡45∘\sin 765^\circ = \sin 45^\circ.

Step 2 — Locate the quadrant.

45∘45^\circ lies in Quadrant I, where sine is positive — so no sign change is needed.

Step 3 — Recall the standard value. …

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