Calculate-without-using-caculator-a-2-2-sin10-2sin35-sec5-2-cos40-sin5-b-sin6-sin42-sin66-sin78-

Question Number 117046 by 1549442205PVT last updated on 09/Oct/20

Calculate without using caculator :

a) - 2 \sqrt{2} \sin 10 ° (2 \sin 35 ° - \frac{\sec 5 °}{2} - \frac{\cos 40 °}{\sin 5 °})

b) \sin 6 ° - \sin 42 ° - \sin 66 ° + \sin 78 °

Answered by bemath last updated on 09/Oct/20

(a) - 4 \sqrt{2} \sin 35 ° \sin 10 ° + \frac{\sqrt{2} \sin 10 °}{\cos 5 °} + \frac{2 \sqrt{2} \cos 40 ° \sin 10 °}{\sin 5 °} =

2 \sqrt{2} (\cos 45 ° - \cos 25 °) + 2 \sqrt{2} \sin 5 ° + 4 \sqrt{2} \cos 40 ° \cos 5 ° =

2 - 2 \sqrt{2} \cos 25 ° + 2 \sqrt{2} \sin 5 ° + 2 \sqrt{2} (\cos 45 ° + \cos 35 °)

4 - 2 \sqrt{2} \cos 25 ° + 2 \sqrt{2} \sin 5 ° + 2 \sqrt{2} \cos 35 ° =

4 - 2 \sqrt{2} \cos 25 ° + 2 \sqrt{2} \cos 35 ° + 2 \sqrt{2} \sin 5 ° =

4 + 2 \sqrt{2} (\cos 35 ° - \cos 25 °) + 2 \sqrt{2} \sin 5 ° =

4 + 2 \sqrt{2} (- 2 \sin 30 ° \sin 5 °) + 2 \sqrt{2} \sin 5 °

4 - 2 \sqrt{2} \sin 5 ° + 2 \sqrt{2} \sin 5 ° = 4

Answered by bobhans last updated on 09/Oct/20

(b) (\sin 78 ° - \sin 66 °) - (\sin 42 ° - \sin 6 °) =

2 \cos 72 ° \sin 6 ° - (2 \cos 24 ° \sin 18 °) =

2 \sin 18 ° \sin 6 ° - 2 \sin 66 ° \sin 18 ° =

2 \sin 18 ° (\sin 6 ° - \sin 66 °) =

2 \sin 18 ° (- 2 \cos 36 ° \sin 30 °) =

- 2 \sin 18 ° \cos 36 ° =

- 2 (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) = - \frac{1}{2}

\Rightarrow \sin 18 ° = \frac{\sqrt{5} - 1}{4}

\Rightarrow \cos 36 ° = \frac{\sqrt{5} + 1}{4}

Answered by 1549442205PVT last updated on 09/Oct/20

Thank Sirs .

Solution :

a) A = - 2 \sqrt{2} (2 \sin 10 \sin 35 - \frac{\sin 10}{2 \cos 5} - \frac{\sin 10 \cos 40}{\sin 5})

= - 2 \sqrt{2} (2 \sin 10 \sin 35 - \sin 5 - 2 \cos 5 \cos 40)

- 2 \sqrt{2} [\cos 25 - \cos 45 - \sin 5 - (\cos 45 + \cos 35)]

= - 2 \sqrt{2} [(\cos 25 - \cos 35) - \sin 5 - 2 \cos 45]

= - 2 \sqrt{2} (2 \sin 30 \sin 5 - \sin 5 - \sqrt{2})

= - 2 \sqrt{2} (2 . \frac{1}{2} \sin 5 - \sin 5 - \sqrt{2})

= - 2 \sqrt{2} (- \sqrt{2}) = 4

b) B = \sin 6 ° - \sin 42 ° - \sin 66 ° + \sin 78 °

= (\sin 6 - \sin 66) + (\sin 78 - \sin 42)

= 2 \cos 36 \sin (- 30) + (2 \cos 60 \sin 18

= - 2 . \frac{1}{2} \cos 36 + 2 \frac{1}{2} \sin 18

= \sin 18 - co 36 = \sin 18 - \sin 54

= - 2 \cos 36 \sin 18 = \frac{- 2 \cos 36 \sin 18 \cos 18}{\cos 18}

= \frac{- \cos 36 \sin 36}{\sin 72} = - \frac{1}{2} \frac{\sin 72}{\sin 72} = - \frac{1}{2}