Question

Simplify the following expression: $\sin {40^\circ} - \cos {70^\circ}$
A) $\sqrt 3 \sin {80^\circ}$
B) $\sqrt 3 \cos {80^\circ}$
C) $\dfrac{1}{2}\cos {80^\circ}$
D) $\dfrac{{\sqrt 3 }}{2}\sin {80^\circ}$

Answer 1

Hint: According to given in the question we have to simplify the given expression $\sin {40^\circ} - \cos {70^\circ}$ so, first of all we have to convert the term $\cos {70^\circ}$ in the form of $\sin {20^\circ}$ with the help of the formula as given below:

Formula used:
$\cos ({90^\circ} - \theta ) = \sin \theta ...............(1)$
After converting the term of the given expression, which is $\cos {70^\circ}$ with the help of the formula (1) to find the simplified form.
Now, as we have obtained both of the term in the form of sin so, we have to use the formula to simplify the obtained expression as given below:

$\sin A - \operatorname{Sin} B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)..........................(2)$
Hence, by applying the formula (2) we can simplify the given trigonometric expression and at the end we have to convert $\cos $ into $\sin $ with the help of the formula given below:
$\sin ({90^\circ} - \theta ) = \cos \theta ...............(3)$
$\cos {30^\circ} = \dfrac{{\sqrt 3 }}{2}......................(a)$

Complete step by step answer:
Step 1: First of all we have to convert the term $\cos {70^\circ}$ in the form of $\sin {20^\circ}$ with the help of the formula (1) as mentioned in the solution hint.
$
   = \sin {40^\circ} - \cos ({90^\circ} - {20^\circ}) \\
   = \sin {40^\circ} - \sin {20^\circ} \\
 $
Step 2: Now, we have to apply the formula (2) to simplify the expression as obtained in step 1.
$ \Rightarrow \sin {40^\circ} - \sin {20^\circ} = 2\cos \left( {\dfrac{{{{40}^\circ} + {{20}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{40}^\circ} - {{20}^\circ}}}{2}} \right)$
Step 3: On solving the obtained expression in step 2.
$
   \Rightarrow \sin {40^\circ} - \sin {20^\circ} = 2\cos \left( {\dfrac{{{{60}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{20}^\circ}}}{2}} \right) \\
   \Rightarrow \sin {40^\circ} - \sin {20^\circ} = 2\cos {30^\circ}\cos {10^\circ} \\
 $
Step 3: Now, we have to use the formula (a) to find the value of $\sin {30^\circ}$ and formula (3) as mentioned in the solution hint to convert $\cos $ into $\sin $
$
   \Rightarrow \sin {40^\circ} - \sin {20^\circ} = 2\cos {30^\circ}\cos {10^\circ} \\
  \Rightarrow \sin {40^\circ} - \sin {20^\circ} = 2 \times \dfrac{{\sqrt 3 }}{2} \times \cos ({90^\circ} - {80^\circ}) \\
   \Rightarrow \sin {40^\circ} - \sin {20^\circ} = \sqrt 3 \sin {80^\circ} \\
 $
Hence, with the help of the formula (1), (2), (3) and (a) we have obtained the value of the given trigonometric expression: $\sin {40^\circ} - \cos {70^\circ} = \sqrt 3 \sin {80^\circ}$.

Therefore option (A) is correct.

Note:
To simplify the given trigonometric expression it is necessary to convert $\cos $ into $\sin $ with the help of $\cos ({90^\circ} - \theta ) = \sin \theta $ to make both of the terms in the form of sin.
While solving the given trigonometric expression, always remember the positive and negative signs used in formula and expression.