Question

If \[A = {130^ \circ }\] and \[x = \sin A + \cos A\] then
A. \[x > 0\]
B. \[x < 0\]
C. \[x = 0\]
D. \[x \leqslant 0\]

Answer 1

Hint: In this question, we need to find the condition for which \[x = \sin A + \cos A\] if \[A = {130^ \circ }\]. For this, we need to put \[A = {130^ \circ }\] in the equation \[x = \sin A + \cos A\]. We will use the following trigonometric identities for the simplification.

Formula used: The following trigonometric identities are useful for solving this question.
\[\cos \left( {90 - \theta } \right) = \sin \theta \]
\[sinA - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)\]
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
\[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\]

Complete step-by-step answer:
We know that \[x = \sin A + \cos A\]
Put \[A = {130^ \circ }\] in the equation \[x = \sin A + \cos A\].
So, we get
\[x = \sin \left( {{{130}^ \circ }} \right) + \cos \left( {{{130}^ \circ }} \right)\]
Let us simplify this.
\[x = \sin \left( {{{180}^ \circ } - {{50}^ \circ }} \right) + \cos \left( {{{180}^ \circ } - {{50}^ \circ }} \right)\]
But we know that \[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\] and \[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\]
Thus, we get
\[x = \sin \left( {{{180}^ \circ }} \right)\cos \left( {{{50}^ \circ }} \right) - \cos \left( {{{180}^ \circ }} \right)\sin \left( {{{50}^ \circ }} \right) + \cos \left( {{{180}^ \circ }} \right)\cos \left( {{{50}^ \circ }} \right) + \sin \left( {{{180}^ \circ }} \right)\sin \left( {{{50}^ \circ }} \right)\]
But \[\sin \left( {{{180}^ \circ }} \right) = 0\] and \[\cos \left( {{{180}^ \circ }} \right) = - 1\]
So, we get
\[x = \left( 0 \right)\cos \left( {{{50}^ \circ }} \right) - \left( { - 1} \right)\sin \left( {{{50}^ \circ }} \right) + \left( { - 1} \right)\cos \left( {{{50}^ \circ }} \right) + \left( 0 \right)\sin \left( {{{50}^ \circ }} \right)\]
\[x = \sin \left( {{{50}^ \circ }} \right) - \cos \left( {{{50}^ \circ }} \right)\]
\[x = \sin \left( {{{50}^ \circ }} \right) - \cos \left( {{{90}^ \circ } - {{40}^ \circ }} \right)\]
But \[\cos \left( {90 - \theta } \right) = \sin \theta \]
Thus, we get
\[x = \sin \left( {{{50}^ \circ }} \right) - \sin \left( {{{40}^ \circ }} \right)\]
By applying the trigonometric identity such as \[sinA - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)\], we get
\[x = 2\sin \left( {\dfrac{{{{50}^ \circ } - {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{50}^ \circ } + {{40}^ \circ }}}{2}} \right)\]
\[x = 2\sin \left( {\dfrac{{{{10}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{90}^ \circ }}}{2}} \right)\]
By simplifying, we get
\[x = 2\sin \left( {{5^ \circ }} \right)\cos \left( {{{45}^ \circ }} \right)\]
Here, we can say that the values of \[\sin \left( {{5^ \circ }} \right)\] and \[\cos \left( {{{45}^ \circ }} \right)\] both are positive.
That means \[x\] is also positive.
Mathematically, we can say that \[x > 0\]
Therefore, the correct option is (A).

Additional information : Trigonometric identities are equalities in trigonometry that are valid for any value of the occurring variables at which both halves of the equation are specified. In simple words, the trigonometric identities are equalities using trigonometric functions that remain true for any value of the variables involved, hence defining both halves of the equality. Hence, Sin, cos, and tan are the three primary trigonometric ratios whereas sec, cosec, and cot are the secondary trigonometric ratios. Also, all the trigonometric identities are associated with the right-angled triangle.

Note: Here, students generally make mistakes in writing the trigonometric identities. They may get confused with the sign in case of writing the formulas of \[\sin \left( {A - B} \right)\] and \[\cos \left( {A - B} \right)\]. Our aim is to reduce the given expression in term of all sine terms to get the required result.