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Question: How do you evaluate \(\sin \left( \arccos .5+\arcsin .6 \right)\) ? (a) Using trigonometric angle...

How do you evaluate sin⁡(arccos⁡.5+arcsin⁡.6)\sin \left( \arccos .5+\arcsin .6 \right) ?
(a) Using trigonometric angle identities
(b) Using linear formulas
(c) a and b both
(d) none of the above

Explanation

Solution

In this problem we are trying to evaluate sin⁡(arccos⁡.5+arcsin⁡.6)\sin \left( \arccos .5+\arcsin .6 \right). We will start by considering the value of arccos⁡(.5)=x\arccos (.5)=x. From which we can reach the value of x in terms of arcsin. Next, we are to use the formula of sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right) . Then by analyzing the properties of trigonometric inverse functions we can conclude that the value we will need.

Complete step by step solution:
According to the problem, we are to evaluate the value of sin⁡(arccos⁡.5+arcsin⁡.6)\sin \left( \arccos .5+\arcsin .6 \right).
Now, the arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. For every trigonometry function, there is an inverse function that works in reverse. These inverse functions have the same name but with 'arc' in front. So the inverse of sin is arcsin. When we see "arcsin A", we understand it as "the angle whose sin is A".
Let us consider that,
arccos⁡(.5)=x\arccos (.5)=x which implies that, cos⁡x=.5=12\cos x=.5=\dfrac{1}{2} .
Now, using trigonometric identities, sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 ,
sin⁡x=1−cos⁡2x\sin x=\sqrt{1-{{\cos }^{2}}x}
As, we are given, cos⁡x=12\cos x=\dfrac{1}{2},
sin⁡x=1−(12)2\sin x=\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}
Simplifying,
⇒1−14\Rightarrow \sqrt{1-\dfrac{1}{4}}
More simplifying,
We get, sin⁡x=32\sin x=\dfrac{\sqrt{3}}{2}
So, x=arcsin⁡32x=\arcsin \dfrac{\sqrt{3}}{2}
Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.
Now, as per to the given question,
sin⁡(arccos⁡.5+arcsin⁡.6)\sin \left( \arccos .5+\arcsin .6 \right)
⇒sin⁡(sin−132+sin⁡−135)\Rightarrow \sin \left( si{{n}^{-1}}\dfrac{\sqrt{3}}{2}+{{\sin }^{-1}}\dfrac{3}{5} \right)
As, said in the trigonometric identities, sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right),
Putting the values, we have,
⇒sin⁡(sin−1(321−925+351−34))\Rightarrow \sin \left( si{{n}^{-1}}\left( \dfrac{\sqrt{3}}{2}\sqrt{1-\dfrac{9}{25}}+\dfrac{3}{5}\sqrt{1-\dfrac{3}{4}} \right) \right)
Doing the simplification,
⇒sin⁡(sin−1(32.45+35.12))\Rightarrow \sin \left( si{{n}^{-1}}\left( \dfrac{\sqrt{3}}{2}.\dfrac{4}{5}+\dfrac{3}{5}.\dfrac{1}{2} \right) \right)
Now, we get,
⇒sin⁡(sin−1(3+4310))\Rightarrow \sin \left( si{{n}^{-1}}\left( \dfrac{3+4\sqrt{3}}{10} \right) \right)
So, using the trigonometric inverse function properties,
⇒3+4310\Rightarrow \dfrac{3+4\sqrt{3}}{10}

So, the correct answer is “Option a”.

Note: Inverse trigonometric functions like arcsin⁡(.6)\arcsin (.6) and arccos⁡(.5)\arccos (.5) are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, anti trigonometric functions or cyclometric functions. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios. Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value.