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

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

Explanation

Solution

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

Complete step-by-step solution:
Given,
sin⁡29∘−cos⁡61∘\sin {29^ \circ } - \cos {61^ \circ }
We can write the value of sin⁡\sin in terms of cos⁡\cos by the complementing function.
sin⁡(90−θ)=cos⁡θ\sin \left( {90 - \theta } \right) = \cos \theta
We can write 2929 as; 29=90−6129 = 90 - 61
Replacing it in the function, we get;
⇒sin⁡(90−61)−cos⁡61\Rightarrow \sin \left( {90 - 61} \right) - \cos 61
Since we know;
sin⁡(90−θ)=cos⁡θ\sin \left( {90 - \theta } \right) = \cos \theta
We get;
sin⁡(90−61)=cos⁡61\sin \left( {90 - 61} \right) = \cos 61
So, substituting it in the expression, we get;
⇒cos⁡61∘−cos⁡61∘\Rightarrow \cos {61^ \circ } - \cos {61^ \circ }
Subtracting the terms, we get;
⇒cos⁡61∘−cos⁡61∘=0\Rightarrow \cos {61^ \circ } - \cos {61^ \circ } = 0
Therefore, we have;
sin⁡29∘−cos⁡61∘=0\sin {29^ \circ } - \cos {61^ \circ } = 0

The value of the given expression is 0.

Additional Information: To prove that sin⁡(90−θ)=cos⁡θ\sin \left( {90 - \theta } \right) = \cos \theta , we can follow the procedure given below:
sin⁡x=osh\sin x = \dfrac{{os}}{h}
Where,
os=os = opposite side to angle xx and
h=h = hypotenuse.
cos⁡(90−x)=ash\cos \left( {90 - x} \right) = \dfrac{{as}}{h}
Where,
as=as = side adjacent to angle 90−x90 - x
h=h = hypotenuse.
We can say that,
Side opposite to angle x$$$$ = side adjacent to angle 90−x90 - x
That implies, the hypotenuse is cancelled. So, we get;
sin⁡x=cos⁡(90−x)\sin x = \cos \left( {90 - x} \right)
Similarly, we can also prove that:
cos⁡x=sin⁡(90−x)\cos x = \sin \left( {90 - x} \right)
The basic three trigonometric identities are sine, cosine and tangent which are short formed into sin⁡,cos⁡\sin ,\cos and tan⁡\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⁡\cos in terms of sin⁡\sin by using the same complimenting function.
Given,
sin⁡29∘−cos⁡61∘\sin {29^ \circ } - \cos {61^ \circ }
We can write the value of cos⁡\cos in terms of sin⁡\sin by the complementing function.
cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta
We can write 6161 as; 61=90−2961 = 90 - 29
Replacing it in the function, we get;
⇒sin⁡29−cos⁡(90−29)\Rightarrow \sin 29 - \cos \left( {90 - 29} \right)
Since we know;
cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta
We get;
cos⁡(90−29)=sin⁡29\cos \left( {90 - 29} \right) = \sin 29
So, substituting it in the expression, we get;
⇒sin⁡29∘−sin⁡29∘\Rightarrow \sin {29^ \circ } - \sin {29^ \circ }
Subtracting the terms, we get;
⇒sin⁡29∘−sin⁡29∘=0\Rightarrow \sin {29^ \circ } - \sin {29^ \circ } = 0
Therefore, we have;
⇒sin⁡29∘−sin⁡29∘=0\Rightarrow \sin {29^ \circ } - \sin {29^ \circ } = 0