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Question: What is the value of \[\cos \left( {2{{\cos }^{ - 1}}0.8} \right)\]? A.\[0.81\] B.\[0.56\] C.\...

What is the value of cos(2cos10.8)\cos \left( {2{{\cos }^{ - 1}}0.8} \right)?
A.0.810.81
B.0.560.56
C.0.480.48
D.0.280.28

Explanation

Solution

Hint First, we will first assume 2cos10.8=θ2{\cos ^{ - 1}}0.8 = \theta and then use this value in the property of trigonometric functions, 1+cosθ=2cos2(θ2)1 + \cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right). Then we will substitute the obtained value of θ\theta in the given equation to find the required answer.

Complete step-by-step answer:
We are given cos(2cos10.8)\cos \left( {2{{\cos }^{ - 1}}0.8} \right).
Let us assume that 2cos10.8=θ2{\cos ^{ - 1}}0.8 = \theta .
Dividing the assumed equation by 2 on both sides, we get

2cos10.82=θ2 cos10.8=θ2  \Rightarrow \dfrac{{2{{\cos }^{ - 1}}0.8}}{2} = \dfrac{\theta }{2} \\\ \Rightarrow {\cos ^{ - 1}}0.8 = \dfrac{\theta }{2} \\\

Taking cos\cos on both sides in the above equation, we get
cos(cos10.8)=cosθ2\Rightarrow \cos \left( {{{\cos }^{ - 1}}0.8} \right) = \cos \dfrac{\theta }{2}
Using the inverse property of trigonometric functions, cos(cos1x)=x\cos \left( {{{\cos }^{ - 1}}x} \right) = x in the above equation, we get

0.8=cosθ2 cosθ2=0.8  \Rightarrow 0.8 = \cos \dfrac{\theta }{2} \\\ \Rightarrow \cos \dfrac{\theta }{2} = 0.8 \\\

We know the property of trigonometric functions, 1+cosθ=2cos2(θ2)1 + \cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right).
Substituting the value of cosθ2\cos \dfrac{\theta }{2} in the property of trigonometry, we get

1+cosθ=2(0.8)2 1+cosθ=2(0.64) 1+cosθ=1.28  \Rightarrow 1 + \cos \theta = 2{\left( {0.8} \right)^2} \\\ \Rightarrow 1 + \cos \theta = 2\left( {0.64} \right) \\\ \Rightarrow 1 + \cos \theta = 1.28 \\\

Subtracting the above equation by 1 on both sides, we get

1+cosθ1=1.281 cosθ=0.28  \Rightarrow 1 + \cos \theta - 1 = 1.28 - 1 \\\ \Rightarrow \cos \theta = 0.28 \\\

Taking cos1{\cos ^{ - 1}} on both sides and using the inverse property again in the above equation, we get

cos1(cosθ)=cos10.28 θ=cos10.28  \Rightarrow {\cos ^{ - 1}}\left( {\cos \theta } \right) = {\cos ^{ - 1}}0.28 \\\ \Rightarrow \theta = {\cos ^{ - 1}}0.28 \\\

Substituting the above value of θ\theta in the given equation, we get

cos(2cos10.8) =cos(θ) =cos(cos10.28) =0.28  \Rightarrow \cos \left( {2{{\cos }^{ - 1}}0.8} \right) \\\ = \cos \left( \theta \right) \\\ = \cos \left( {{{\cos }^{ - 1}}0.28} \right) \\\ = 0.28 \\\

So, the answer is 0.280.28.
Hence, option D is correct.

Note In solving this question, we should know the basic properties of trigonometric functions, like 1+cosθ=2cos2(θ2)1 + \cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right) and cos(cos1x)=x\cos \left( {{{\cos }^{ - 1}}x} \right) = x. If students are familiar with the properties, then these types of questions are simple. Students have to be really careful while solving to avoid the calculation mistakes.