If A = 30°, then prove that :
2 cos2 A - 1 = 1 - 2 sin2 A
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L.H.S.=2cos2A−1=2cos230°−1=2×(32)2−1=2×34−1=32−22=3−22=12\text{L.H.S.} = 2 \text{cos}^2 A - 1\\[1em] = 2 \text{cos}^2 30° - 1\\[1em] = 2 \times \Big(\dfrac{\sqrt3}{2}\Big)^2 - 1\\[1em] = 2 \times \dfrac{3}{4} - 1\\[1em] = \dfrac{3}{2} - \dfrac{2}{2}\\[1em] = \dfrac{3 - 2}{2}\\[1em] = \dfrac{1}{2}L.H.S.=2cos2A−1=2cos230°−1=2×(23)2−1=2×43−1=23−22=23−2=21
R.H.S.=1−2sin2A=1−2sin230°=1−2×(12)2=1−2×14=22−12=2−12=12\text{R.H.S.} = 1 - 2 \text{sin}^2 A \\[1em] = 1 - 2 \text{sin}^2 30°\\[1em] = 1 - 2 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = 1 - 2 \times \dfrac{1}{4}\\[1em] = \dfrac{2}{2} - \dfrac{1}{2}\\[1em] = \dfrac{2 - 1}{2}\\[1em] = \dfrac{1}{2}R.H.S.=1−2sin2A=1−2sin230°=1−2×(21)2=1−2×41=22−21=22−1=21
∴ L.H.S. = R.H.S.
Hence, 2 cos2 A - 1 = 1 - 2 sin2 A.
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sin 2 A=2 sin A cos A=2 tan A1+tan2 A\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}sin 2 A=2 sin A cos A=1+tan2 A2 tan A
cos 2 A=cos2 A−sin2 A=1−tan2 A1+tan2 A\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}cos 2 A=cos2 A−sin2 A=1+tan2 A1−tan2 A
sin 3A = 3 sin A - 4 sin3 A
If A = B = 45°, show that :
sin (A - B) = sin A cos B - cos A sin B
