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Question: How do I find the value of \(\sin (\dfrac{{5\pi }}{6})\)?...

How do I find the value of sin⁡(5π6)\sin (\dfrac{{5\pi }}{6})?

Explanation

Solution

This sum can be solved using two methods. It is upto the students which one he/she prefers. First method is using the formulasin⁡(180−θ)=sin⁡θ\sin (180 - \theta ) = \sin \theta . This formula is derived using the graphical properties of the sin function. Second method is to use the formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B. Both methods would give the same answer, only difference is that the second method is a bit longer compared to the first one.

Complete step by step solution:
Since there are two methods, let's start with the shorter method.
We know that the value of πis180∘\pi is{180^ \circ }, therefore 5π6=π−π6\dfrac{{5\pi }}{6} = \pi - \dfrac{\pi }{6}.
We know that sin⁡(180−θ)=sin⁡θ\sin (180 - \theta ) = \sin \theta , from above we can rearrange the value of 5π6\dfrac{{5\pi }}{6} in the given question as sin⁡(5π6)=sin⁡(π−π6)\sin (\dfrac{{5\pi }}{6}) = \sin (\pi - \dfrac{\pi }{6})
Using the property of sin function, we can say that the above equation is equivalent to sin⁡(π6)\sin (\dfrac{\pi }{6})
We know the value of sin⁡(30∘)\sin ({30^ \circ }).

Therefore the answer is 0.50.5.

Second Method:
We can say that sin⁡(5π6)=sin⁡(150)\sin (\dfrac{{5\pi }}{6}) = \sin (150) since the value of πis180∘\pi is{180^ \circ }.
Using the formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B, we can input A=90∘,B=60∘A = {90^ \circ },B = {60^ \circ } . Substituting the value of A & B in the formula
sin⁡(150)=sin⁡90∘cos⁡60∘+cos⁡90∘sin⁡60∘......(1)\sin (150) = \sin {90^ \circ }\cos {60^ \circ } + \cos {90^ \circ }\sin {60^ \circ }......(1)
Since the value of cos⁡90∘\cos {90^ \circ }is 00,we can say that
sin⁡(150)=sin⁡90∘cos⁡60∘...........(2)\sin (150) = \sin {90^ \circ }\cos {60^ \circ }...........(2)
Also we know that sin⁡90∘=1&cos⁡(60∘)=12\sin {90^ \circ } = 1\& \cos ({60^ \circ }) = \dfrac{1}{2},

Therefore value of sin⁡(150∘)\sin ({150^ \circ })is 1×12=0.51 \times \dfrac{1}{2} = 0.5

Note:
Though there are two methods for this type of sums, it is always preferable to use the first method since it is shorter and much easier to comprehend. Second method is preferable when the values of A & B are given and the student has to find the answer to sin⁡(A+B)\sin (A + B). Also the student should memorize the graphical properties of trigonometric functions for angles greater than 90∘,180∘,270∘{90^ \circ },{180^ \circ },{270^ \circ } as there may be a change in the behavior, for example cos⁡(90∘+θ)=−cos⁡(θ)\cos ({90^ \circ } + \theta ) = - \cos (\theta )and so on for other trigonometric functions.