Question

Solve the given question in a detail manner:
Evaluate \[\sin {29^ \circ } - \cos {61^ \circ }\]

Answer 1

Hint: Given terms are written. The complementing function, i.e., \[\sin \left( {90 - \theta } \right) = \cos \theta \] is used. Replace the \[\sin \] in terms of \[\cos \] by the complementing function. After replacing the function, we solve the expression by trigonometric operations.

Complete step-by-step solution:
Given,
\[\sin {29^ \circ } - \cos {61^ \circ }\]
We can write the value of \[\sin \] in terms of \[\cos \] by the complementing function.
\[\sin \left( {90 - \theta } \right) = \cos \theta \]
We can write \[29\] as; \[29 = 90 - 61\]
Replacing it in the function, we get;
\[ \Rightarrow \sin \left( {90 - 61} \right) - \cos 61\]
Since we know;
\[\sin \left( {90 - \theta } \right) = \cos \theta \]
We get;
\[\sin \left( {90 - 61} \right) = \cos 61\]
So, substituting it in the expression, we get;
\[ \Rightarrow \cos {61^ \circ } - \cos {61^ \circ }\]
Subtracting the terms, we get;
\[ \Rightarrow \cos {61^ \circ } - \cos {61^ \circ } = 0\]
Therefore, we have;
\[\sin {29^ \circ } - \cos {61^ \circ } = 0\]

The value of the given expression is 0.

Additional Information: To prove that \[\sin \left( {90 - \theta } \right) = \cos \theta \], we can follow the procedure given below:
\[\sin x = \dfrac{{os}}{h}\]
Where,
\[os = \]opposite side to angle \[x\] and
\[h = \]hypotenuse.
\[\cos \left( {90 - x} \right) = \dfrac{{as}}{h}\]
Where,
\[as = \]side adjacent to angle \[90 - x\]
\[h = \]hypotenuse.
We can say that,
Side opposite to angle \[x\]\[ = \] side adjacent to angle \[90 - x\]
That implies, the hypotenuse is cancelled. So, we get;
\[\sin x = \cos \left( {90 - x} \right)\]
Similarly, we can also prove that:
\[\cos x = \sin \left( {90 - x} \right)\]
The basic three trigonometric identities are sine, cosine and tangent which are short formed into \[\sin ,\cos \] and \[\tan \] respectively. Here, we can write one function of the trigonometric identity into the other two terms or the six present identities however we want by simple operations.

Note: This sum can also be done using another method that is replacing \[\cos \] in terms of \[\sin \] by using the same complimenting function.
Given,
\[\sin {29^ \circ } - \cos {61^ \circ }\]
We can write the value of \[\cos \] in terms of \[\sin \] by the complementing function.
\[\cos \left( {90 - \theta } \right) = \sin \theta \]
We can write \[61\] as; \[61 = 90 - 29\]
Replacing it in the function, we get;
\[ \Rightarrow \sin 29 - \cos \left( {90 - 29} \right)\]
Since we know;
\[\cos \left( {90 - \theta } \right) = \sin \theta \]
We get;
\[\cos \left( {90 - 29} \right) = \sin 29\]
So, substituting it in the expression, we get;
\[ \Rightarrow \sin {29^ \circ } - \sin {29^ \circ }\]
Subtracting the terms, we get;
\[ \Rightarrow \sin {29^ \circ } - \sin {29^ \circ } = 0\]
Therefore, we have;
\[ \Rightarrow \sin {29^ \circ } - \sin {29^ \circ } = 0\]