Consider an isosceles triangle ABC, where AB=AC=1, M is the midpoint of BC, and <BAM=<CAM=x. Use trigonometry to find an expression for BM and by finding BC^2, show that cos2x = 1 - 2(sinx)^2.
Consider the right-angled triangle BAM. We have the hypotenuse and an angle, and want to find the opposite side. Therefore, using SOHCAHTOA, sinx = BM/1, hence BM = sinx. Because M is the midpoint of BC, BC = 2BM = 2sinx. Hence, BC2 = 4(sinx)2Looking at triangle ABC, we have an expression for all sides (AB=AC=1, BC=2sinx) and we have the angle at A (2x). Therefore, we can use the cosine rule to find an expression for BC2.a2 = b2 + c2 - 2bc(cosA).BC2 = 4(sinx)2 = 12 + 12 - 2(1)(1)(cos2x)4(sinx)2 = 2 - 2cos2x2(sinx)2 = 1 - cos2xcos2x = 1 - 2(sinx)2
BS
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