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Question: The value of \[\operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0}\]is...

The value of Sin⁡500−Sin⁡700+Sin⁡100\operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0}is equal to:
A. 11
B. 00
C. 12\dfrac{1}{2}
D. 22

Explanation

Solution

In order to determine the given function, Sin⁡500−Sin⁡700+Sin⁡100\operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0}. Here, the given equation is in the form of degree and we know the trigonometric identity,
Sin⁡(C)−Sin⁡(D)=2cos⁡(C+D)2sin⁡(C−D)2\operatorname{Sin} (C) - \operatorname{Sin} (D) = 2\cos \dfrac{{(C + D)}}{2}\sin \dfrac{{(C - D)}}{2}
So, by using the above equation, we will substitute for the given question to achieve our solution.
When trigonometric functions are used in an expression or equation, trigonometric Identities come in handy. Identity inequalities are inequalities that hold for any value on both sides of an equation. These identities are geometrically defined as functions of one or more angles.

Complete step by step answer:
We are given the function,
Sin⁡500−Sin⁡700+Sin⁡100\operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0}
Now, by using the formula which we have mentioned in the above, which is:
Sin⁡(C)−Sin⁡(D)=2Cos⁡(C+D)2Sin⁡(C−D)2\operatorname{Sin} (C) - \operatorname{Sin} (D) = 2\operatorname{Cos} \dfrac{{(C + D)}}{2}\operatorname{Sin} \dfrac{{(C - D)}}{2}
we will take C=50,C = 50, and D=70D = 70
So, we will get;

=Sin⁡500−Sin⁡700+Sin⁡100 =2(Cos⁡((500+700)2)Sin⁡((500−700)2))+Sin⁡100 By simplify the degree of the function, we get=2(Cos⁡(12002)Sin⁡(−2002))+Sin⁡100 =2(Cos⁡(600)Sin⁡(−100))+Sin⁡100  = \operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0} \\\ = 2\left( {\operatorname{Cos} \left( {\dfrac{{({{50}^0} + {{70}^0})}}{2}} \right)\operatorname{Sin} \left( {\dfrac{{({{50}^0} - {{70}^0})}}{2}} \right)} \right) + \operatorname{Sin} {10^0} \\\ By\ simplify\ the\ degree\ of\ the\ function,\ we\ get = 2\left( {\operatorname{Cos} \left( {\dfrac{{{{120}^0}}}{2}} \right)\operatorname{Sin} \left( {\dfrac{{ - {{20}^0}}}{2}} \right)} \right) + \operatorname{Sin} {10^0} \\\ = 2(\operatorname{Cos} ({60^0})\operatorname{Sin} ( - {10^0})) + \operatorname{Sin} {10^0} \\\

From the above equation, we know that,
⇒Cos⁡(600)=12\Rightarrow \operatorname{Cos} ({60^0}) = \dfrac{1}{2},
and we also know that,
⇒Sin⁡(−θ)=−Sin⁡θ\Rightarrow \operatorname{Sin} ( - \theta ) = - \operatorname{Sin} \theta
And by substituting the above two formulas, then we will get;

=−2(12)Sin⁡(100)+Sin⁡100 =−Sin⁡(100)+Sin⁡(100) =0  = - 2\left( {\dfrac{1}{2}} \right)\operatorname{Sin} ({10^0}) + \operatorname{Sin} {10^0} \\\ = - \operatorname{Sin} ({10^0}) + \operatorname{Sin} ({10^0}) \\\ = 0 \\\

Therefore, the answer for the given question is:
Sin⁡500−Sin⁡700+Sin⁡100\operatorname{Sin} {50^0} - \operatorname{Sin} {70^0} + \operatorname{Sin} {10^0}is equal to 00

So, the correct answer is “Option B”.

Note: In the given question we have used two formulas, which are very important to achieve our solution and also, we also have used Trigonometric Table Values which are very-very important to know the values, without we will not be able to come to our solution.
We need to solve that given problem by using trigonometric identity and substitution methods.
So, the trigonometric table is nothing but the ratios that are inter-related to each of Sine, Cosine, Tangent, Cosecant, Secant, Cotangent, which are in short Sin, Cos, Tan, Cosec, Sec, and Cot.
This ratio table commonly consist of standard trigonometric values such as 0∘,30∘,45∘,60∘{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ } and 90∘{90^ \circ }So, these are very important to know, and one need to remember these standard angles.