Prove that :- 1) Sin (45 + a) * Sin (45 - a) = 1/2 Cos 2a 2) - Maths - Trigonometric Functions

Question

Prove that :-
1) Sin (45 ° + a) * Sin (45
° - a) = 1/2 Cos 2a
2) (2Cos2a + 1) / (2Cos2a - 1) = tan (60° + a) * tan
(60* - a)
3) 4SinA*Sin(A +π/3)*Sin(A + 2 π/3) =
Sin3A
4) Sin10° Sin50 ° Sin60
° Sin70 ° = ( √3) /
16
5) Cos20° Cos30 ° Cos40
° Cos80 ° = (
√3) / 16

Ankush Jain · Accepted Answer

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1.

To prove: sin (45° + a) × sin (45° - a) = 1/2 cos 2a

Proof:

LHS = sin (45° + a) × sin (45° - a)

$= (\sin 45^{\circ} \cos a + \cos 45^{\circ} \sin a) \times (\sin 45^{\circ} \cos a - \cos 45^{\circ} \sin a) = (\frac{1}{\sqrt{2}} \cos a + \frac{1}{\sqrt{2}} \sin a) \times (\frac{1}{\sqrt{2}} \cos a - \frac{1}{\sqrt{2}} \sin a) = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times (\cos a + \sin a) \times (\cos a - \sin a) = \frac{1}{2} (\cos^{2} a - \sin^{2} a) = \frac{1}{2} \cos 2 a [u \sin g : \cos 2 a = \cos^{2} a - \sin^{2} a] = R H S [H e n c e p r o v e d]$

4.

L.H.S. = sin 10° sin 50° sin 60° sin 70°

$= \sin 10 ° \sin 50 ° (\frac{\sqrt{3}}{2}) \sin 70 ° [∵ \sin 60 ° = \frac{\sqrt{3}}{2}] = \frac{\sqrt{3}}{2} [\frac{1}{2} (2 \sin 10 ° \sin 50 °)] \sin 70 ° = \frac{\sqrt{3}}{4} [\cos (10 ° - 50 °) - \cos (10 ° + 50 °)] \sin 70 ° = \frac{\sqrt{3}}{4} [\cos (- 40 °) - \cos 60 °] \sin 70 ° = \frac{\sqrt{3}}{4} [\cos 40 ° - \frac{1}{2}] \sin 70 ° [∵ \cos 60 ° = \frac{1}{2}] = \frac{\sqrt{3}}{4} [\cos 40 ° \sin 70 °] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{4 \times 2} [2 \sin 70 ° \cos 40 °] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{8} [\sin (70 ° + 40 °) + \sin (70 ° - 40 °)] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{8} [\sin 110 ° + \sin 30 ° - \sin 70 °] = \frac{\sqrt{3}}{8} [\sin (180 ° - 70 °) + \frac{1}{2} - \sin 70 °] = \frac{\sqrt{3}}{8} [\sin 70 ° + \frac{1}{2} - \sin 70 °] [∵ \sin (180 ° - θ) = \sin θ] = \frac{\sqrt{3}}{8 \times 2} = \frac{\sqrt{3}}{16} = R.H.S$

5.

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2

Prove that :- 1) Sin (45 ° + a) * Sin (45 ° - a) = 1/2 Cos 2a 2) (2Cos2a + 1) / (2Cos2a - 1) = tan (60° + a) * tan (60* - a) 3) 4SinA*Sin(A +π/3)*Sin(A + 2 π/3) = Sin3A 4) Sin10°.Sin50 °.Sin60 °.Sin70 ° = ( √3) / 16 5) Cos20°.Cos30 °.Cos40 °.Cos80 ° = ( √3) / 16

Prove that :-

1) Sin (45 ° + a) * Sin (45 ° - a) = 1/2 Cos 2a

2) (2Cos2a + 1) / (2Cos2a - 1) = tan (60° + a) * tan (60* - a)

3) 4SinASin(A +π/3)Sin(A + 2 π/3) = Sin3A

4) Sin10°.Sin50 °.Sin60 °.Sin70 ° = ( √3) / 16

5) Cos20°.Cos30 °.Cos40 °.Cos80 ° = ( √3) / 16