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Question: Solve the equation : \(2\cos 22{\dfrac{1}{2}^0}\sin 22{\dfrac{1}{2}^0}\,\,\) by using trigonometric ...

Solve the equation : 2cos22120sin221202\cos 22{\dfrac{1}{2}^0}\sin 22{\dfrac{1}{2}^0}\,\, by using trigonometric formulas or identities.

Explanation

Solution

Hint - In order to solve this problem use the formula that sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta . Then put the value of angle.

As we know the formula,
sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta ……(i)
And the given equation is 2cos22120sin221202\cos 22{\dfrac{1}{2}^0}\sin 22{\dfrac{1}{2}^0}\,\,
We also know abc=ac+bca\dfrac{b}{c} = \dfrac{{ac + b}}{c} ……(ii)
From (ii) we can say, 2212=22(2)+12=44+12=45222\dfrac{1}{2} = \dfrac{{22(2) + 1}}{2} = \dfrac{{44 + 1}}{2} = \dfrac{{45}}{2} ……(iii)

Therefore the given equation can be written as ,
2cos4520sin45202\cos {\dfrac{{45}}{2}^0}\sin {\dfrac{{45}}{2}^0}\,\,
And from (i) we can say,
2cos4520sin4520=sin2452=sin4502\cos {\dfrac{{45}}{2}^0}\sin {\dfrac{{45}}{2}^0}\,\, = \sin \dfrac{{2\,{\text{x }}45}}{2} = \sin {45^0}
We know the value of sin45 is 12\dfrac{1}{{\sqrt 2 }}.
Therefore from the above equations we can say
2cos22120sin22120=122\cos 22{\dfrac{1}{2}^0}\sin 22{\dfrac{1}{2}^0} = \dfrac{1}{{\sqrt 2 }}.

Hence the answer to this question is 12\dfrac{1}{{\sqrt 2 }}.

Note – In these types of problems of trigonometry we have to use the general formula of trigonometry, after observing which formula can be fit into the given question then solve the equation according to the formula. There is an alternative method to solve this question , it is , if we know the values of the angle given in the equation we can also directly put it and get the actual value of the equation.