What is the value of sin 10° + sin 50° - sin 70° ?

Question

What is the value of sin 10° + sin 50° - sin 70° ? (A) 1 (B) 0 (C) (1/2) (D) 2

Tardigrade Admin · Accepted Answer

The given expression sin 10° + sin 50° - sin 70° can also be written as sin 10° - ( sin 70° - sin 50°) = sin 10° - (2 cos 60° sin 10°) = sin 10° (2 cos 60° - 1) = sin 10°( 1 - 1) = 0

Q. What is the value of $sin 1 0^{\circ} + sin 5 0^{\circ} - sin 7 0^{\circ}$ ?

1

0

$\frac{1}{2}$

2

The given expression $sin 1 0^{\circ} + sin 5 0^{\circ} - sin 7 0^{\circ}$ can also be written as $sin 1 0^{\circ} - (sin 7 0^{\circ} - sin 5 0^{\circ})$
$= sin 1 0^{\circ} - (2 cos 6 0^{\circ} sin 1 0^{\circ})$
$= sin 1 0^{\circ} (2 cos 6 0^{\circ} - 1) = sin 1 0^{\circ} (1 - 1) = 0$

Q. What is the value of sin10∘+sin50∘−sin70∘ ?

1

0

21​

2

The given expression sin10∘+sin50∘−sin70∘ can also be written as sin10∘−(sin70∘−sin50∘) =sin10∘−(2cos60∘sin10∘) =sin10∘(2cos60∘−1)=sin10∘(1−1)=0

Q. What is the value of $sin 1 0^{\circ} + sin 5 0^{\circ} - sin 7 0^{\circ}$ ?

$\frac{1}{2}$

The given expression $sin 1 0^{\circ} + sin 5 0^{\circ} - sin 7 0^{\circ}$ can also be written as $sin 1 0^{\circ} - (sin 7 0^{\circ} - sin 5 0^{\circ})$
$= sin 1 0^{\circ} - (2 cos 6 0^{\circ} sin 1 0^{\circ})$
$= sin 1 0^{\circ} (2 cos 6 0^{\circ} - 1) = sin 1 0^{\circ} (1 - 1) = 0$