Question

find the value of sin3°

Answer 1

Given :  sin3°

To Find : Value of   sin3°

Solution  :

f(x)  = sin(x)

f(x + Δx)  = sin(x + Δx)

x = 0

Δx = 3°  = 3 * π/180  = 0.0523

f'(x)  = cos(x)

f(x + Δx)   = f(x)  + Δx f'(x)

=> sin ( 0 + 3°) = sin (0°)  + ( 0.0523 ) cos(0°)

=> Sin (3°) = 0  +  ( 0.0523 ) (1)

=>  Sin (3°) = 0.0523

value of sin3° = 0.0523

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Answer 2

To Find : the value of sin3°

Step-by-step explanation:

sin 15° = $\frac{\sqrt{6} - \sqrt{2} }{4}$

sin 18° = $\frac{1}{4} (\sqrt{5} - 1)$

sin 72° = $\frac{\sqrt{2} }{4}$ $\sqrt{5+\sqrt{5} }$

sin 75° = $\frac{\sqrt{6}+ \sqrt{2} }{4}$

By using,

    sin (a - b) = sin a . cos b - cos a . sin b

sin (18° - 15°) = sin 18°. cos 15° - cos 18°. sin 15°

    sin 3°   = sin 18°. sin 75° - sin 72°. sin 15°

    sin 3°    = $\frac{1}{4} (\sqrt{5} - 1)$ . $\frac{\sqrt{6}+ \sqrt{2} }{4}$  - $\frac{\sqrt{2} }{4}$ $\sqrt{5+\sqrt{5} }$ .  $\frac{\sqrt{6} - \sqrt{2} }{4}$

    sin 3°    = $\frac{1}{16}$ [ ($\sqrt{5} - 1$) ($\sqrt{6} + \sqrt{2}$) - 2($2\sqrt{3} -2$) ($\sqrt{5+\sqrt{5} }$) ]

    sin 3°    =  $\frac{1}{16}$ [ ($\sqrt{5} - 1$) ($\sqrt{6} + \sqrt{2}$) - $\frac{1}{8}$ ($\sqrt{3} - 1$) ($\sqrt{5+\sqrt{5} }$) ]