If A = 30°, then prove that :
sin 3A = 3 sin A - 4 sin3 A
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L.H.S. = sin 3A
= sin (3 x 30°)
= sin 90°
= 1
R.H.S.=3sin A−4sin3A=3sin 30°−4sin330°=3×12−4×(12)3=32−4×18=32−12=3−12=22=1\text{R.H.S.} = 3 \text{sin A} - 4 \text{sin}^3 A\\[1em] = 3 \text{sin 30°} - 4 \text{sin}^3 30°\\[1em] = 3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3\\[1em] = \dfrac{3}{2} - 4 \times \dfrac{1}{8}\\[1em] = \dfrac{3}{2} - \dfrac{1}{2}\\[1em] = \dfrac{3 - 1}{2}\\[1em] = \dfrac{2}{2}\\[1em] = 1R.H.S.=3sin A−4sin3A=3sin 30°−4sin330°=3×21−4×(21)3=23−4×81=23−21=23−1=22=1
∴ L.H.S. = R.H.S.
Hence, sin 3A = 3 sin A - 4 sin3 A.
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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
2 cos2 A - 1 = 1 - 2 sin2 A
If A = B = 45°, show that :
sin (A - B) = sin A cos B - cos A sin B
cos (A + B) = cos A cos B - sin A sin B
