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Exercise

sin⁡(x−2π)\sin\left(x-\frac{2}{\pi}\right)

Step-by-step Solution

1

Using the sine of a sum formula: sin⁡(α±β)=sin⁡(α)cos⁡(β)±cos⁡(α)sin⁡(β)\sin(\alpha\pm\beta)=\sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta), where angle α\alpha equals xx, and angle β\beta equals −2π-\frac{2}{\pi }

cos⁡(2π)sin⁡(x)−sin⁡(2π)cos⁡(x)\cos\left(\frac{2}{\pi }\right)\sin\left(x\right)-\sin\left(\frac{2}{\pi }\right)\cos\left(x\right)

Final answer to the exercise

cos⁡(2π)sin⁡(x)−sin⁡(2π)cos⁡(x)\cos\left(\frac{2}{\pi }\right)\sin\left(x\right)-\sin\left(\frac{2}{\pi }\right)\cos\left(x\right)

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sin(x−2π )
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