Question

Find the value of \[\sin \,{50^ \circ } - \sin \,{70^ \circ } + \sin \,{10^ \circ }\]
A. \[0\]
B. \[1\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{{\sqrt 2 }}\]

Answer 1

Hint: In this question, we use the \[\sin \,A + \sin \,B\] formula. First, we take \[\sin \,{50^ \circ } + \sin \,{10^ \circ }\]then simplify it with the formula, and then re-insert it into the equation. To solve the remaining problems, we use trigonometry's cosine function.

Formula used:
1. \[\sin \,A + \sin \,B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
2. \[\cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta \]
3. \[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
4. \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]

Complete step-by-step solution:
We are given that \[\sin \,{50^ \circ } - \sin \,{70^ \circ } + \sin \,{10^ \circ }...\left( 1 \right)\]
We are asked to find the value of \[\sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ }\]
Now we know the formula of \[\sin \,A + \sin \,B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
Now we take \[\sin \,{50^ \circ } + \sin \,{10^ \circ }\]where \[A = {50^ \circ },B = {10^ \circ }\]
By the above formula, we have
\[
  \sin {50^ \circ } + \sin {10^ \circ } = 2\sin \left( {\dfrac{{{{50}^ \circ } + {{10}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{50}^ \circ } - {{10}^ \circ }}}{2}} \right) \\
   = 2\sin \left( {\dfrac{{{{60}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{40}^ \circ }}}{2}} \right) \\
   = 2\sin \left( {{{30}^ \circ }} \right)\cos \left( {{{20}^ \circ }} \right)
 \]
Now we know that \[\sin \,{30^ \circ } = \dfrac{1}{2}\]
Therefore,
\[
  \sin {50^ \circ } + \sin {10^ \circ } = 2 \times \dfrac{1}{2}\,\cos \left( {{{20}^ \circ }} \right) \\
   = \cos \left( {{{20}^ \circ }} \right)...\left( 2 \right)
 \]
Now re-insert this value in equation (1) and we get
\[\cos \,{20^ \circ } - \sin \,{70^ \circ }...\left( 3 \right)\]
Now we know that \[\cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta \]
\[
  \cos \,{20^ \circ } = \cos \left( {{{90}^ \circ } - {{70}^ \circ }} \right) \\
   = \sin \,{70^ \circ }
 \]
When we substitute the above value in equation (4), we get
\[\sin \,{70^ \circ } - \sin \,{70^ \circ } = 0\]
Therefore, the value of \[\sin \,{50^ \circ } - \sin \,{70^ \circ } + \sin \,{10^ \circ }\] is \[0\].
Hence, option (A) is the correct answer.

Additional information: Trigonometric identities are used in mathematics when a trigonometric function is involved with an expression or an equation. When using trigonometry in geometry, the identities are involved with the function or the angles. Many trigonometric identities involve the angle of the triangle and the side length of the triangle. The trigonometric identities are only applicable to right-angled triangles. Trigonometric identity is represented with the help of an equation that has trigonometric ratios.

Note: We can also solve this question by an alternate method:
Given that \[\sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ }\]
Now we know that
\[
  \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B \\
  \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
 \]
Now, \[\sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ }\] can be written as
\[
  \sin \left( {{{70}^ \circ }} \right) = \sin {60^ \circ }\cos {10^ \circ } + \cos {60^ \circ }\sin {10^ \circ } \\
  \sin \left( {{{50}^ \circ }} \right) = \sin {60^ \circ }\cos {10^ \circ } - \cos {60^ \circ }\sin {10^ \circ }
 \]
Therefore,
\[
  \sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ } = \sin {60^ \circ }\cos {10^ \circ } - \cos {60^ \circ }\sin {10^ \circ } - \left[ {\sin {{60}^ \circ }\cos {{10}^ \circ } + \cos {{60}^ \circ }\sin {{10}^ \circ }} \right] + \sin {10^ \circ } \\
   = \sin {60^ \circ }\cos {10^ \circ } - \cos {60^ \circ }\sin {10^ \circ } - \sin {60^ \circ }\cos {10^ \circ } - \cos {60^ \circ }\sin {10^ \circ } + \sin {10^ \circ }
 \]
Now, by canceling out the like terms, we get
\[\sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ } = - 2\cos {60^ \circ }\sin {10^ \circ } + \sin {10^ \circ }\]
Now we know that \[\cos {60^ \circ } = \dfrac{1}{2}\]
Therefore,
\[
  \sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ } = - 2 \times \dfrac{1}{2}\,\sin {10^ \circ } + \sin {10^ \circ } \\
   = - \sin {10^ \circ } + \sin {10^ \circ } \\
   = 0
 \]
Hence, the value of \[\sin {50^ \circ } - \sin {70^ \circ } + \sin {10^ \circ }\]is \[0\].