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Trigonometry · real student question

Evaluate cos(sin(3)), with the argument in radians.

Question

Evaluate

cos⁡(sin⁡3)\cos(\sin 3)

where the argument is in radians.

Step-by-step solution

  1. Work from the inside out. This is a composition, not a product: cos⁡(sin⁡3)\cos(\sin 3) means take sin⁡3\sin 3 first, then feed that number into cos⁡\cos. It is not cos⁡3×sin⁡3\cos3\times\sin3.

  2. Settle the angle mode before computing. With no degree symbol, 33 means 33 radians. Since π≈3.14159\pi\approx3.14159, an angle of 33 radians is just short of a half turn — near 171.9∘171.9^\circ, in the second quadrant, where sine is small and positive.

  3. Evaluate the inner function.

    sin⁡3=0.1411200\sin3=0.1411200

    This is small precisely because 33 is close to π\pi, where sine vanishes; indeed sin⁡3≈sin⁡(π−3)=sin⁡(0.14159)≈0.1412\sin3\approx\sin(\pi-3)=\sin(0.14159)\approx0.1412 ✓.

  4. Evaluate the outer function on that result. Note that sin⁡3\sin3 is itself an angle in radians now, and it lies in [−1,1][-1,1] as every sine value must:

    cos⁡(0.1411200)=0.9900590\cos(0.1411200)=0.9900590

    For small θ\theta, cos⁡θ≈1−θ22=1−0.1411222=1−0.00996=0.99004\cos\theta\approx1-\tfrac{\theta^2}{2}=1-\tfrac{0.14112^2}{2}=1-0.00996=0.99004 — matching to four decimals ✓.

  5. State the answer.

    cos⁡(sin⁡3)≈0.99006\cos(\sin3)\approx0.99006

  6. Note the range of the whole composition. Because sin⁡3\sin3 can only lie in [−1,1][-1,1], the outer cosine is evaluated on [−1,1][-1,1], where it is at least cos⁡1=0.5403\cos1=0.5403. So cos⁡(sin⁡x)\cos(\sin x) always lies in [0.5403,1][0.5403,1] for every real xx — it can never be negative. Getting a negative answer would be a sure sign of a mode or bracketing error.

Answer

cos⁡(sin⁡3)=cos⁡(0.14112)≈0.99006\cos(\sin 3)=\cos(0.14112)\approx 0.99006

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