Given that tan⁡x2=1−cos⁡(x)sin⁡(x), deduce that tan⁡π12=2−3 I know tan⁡x2=1−cos⁡(x)sin⁡(x) is true and I can prove it by squaring and taking a square root of the right side then I multiply by 0.50.5. And I will use cos⁡x2=1+cos⁡x2 and sin⁡x2=1−cos⁡x2 But I do not understand the part of deducing.

Question

Given that tan⁡x2=1−cos⁡(x)sin⁡(x), deduce that tan⁡π12=2−3  I know tan⁡x2=1−cos⁡(x)sin⁡(x) is true and I can prove it by squaring and taking a square root of the right side then I multiply by 0.50.5. And I will use  cos⁡x2=1+cos⁡x2  and  sin⁡x2=1−cos⁡x2  But I do not understand the part of deducing.

Madelyn Townsend · Accepted Answer

I suspect this is what you need to solve:  Show that tan⁡(112π)=2−3, given that  tan⁡(x2)=1−cos⁡xsin⁡x  This can be shown quite easily as cos⁡(16π)=123 and sin⁡(16π)=12. (These are well known values and I suspect you do not need to proof this.)

Given that \tan \frac x2=\frac{1-\cos(x)}{sin(x)}, deduce that \tan \frac{\pi}{12}=2−\sqrt3

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