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Question: State true or false for the given trigonometric expression \(\cos 11{}^\circ -\cos 2{}^\circ >0\), ...

State true or false for the given trigonometric expression cos11cos2>0\cos 11{}^\circ -\cos 2{}^\circ >0,
A. True
B. False

Explanation

Solution

Hint: We will first calculate the value of cos11\cos 11{}^\circ and then we will calculate the value of cos2\cos 2{}^\circ . Finally we will subtract the value of cos2\cos 2{}^\circ with cos11\cos 11{}^\circ to get the final result.

Complete step-by-step answer:
It is given that we have to find the relation cos11cos2>0\cos 11{}^\circ -\cos 2{}^\circ >0 is true or false.
This picture and the values below the picture shows cosθ\cos \theta values at different angles.

cosθ\cos \theta
Values
cos0\cos 0{}^\circ =1
cos30\cos 30{}^\circ =32\dfrac{\sqrt{3}}{2}
cos45\cos 45{}^\circ =12\dfrac{1}{\sqrt{2}}
cos60\cos 60{}^\circ =12\dfrac{1}{2}
cos90\cos 90{}^\circ =0
cos180\cos 180{}^\circ =-1

So, from above we get an idea about the trend of cosθ\cos \theta and their values in different angles. Please note that the value of cosθ\cos \theta decreases with the increase in angle from the of cosθ\cos \theta .

In the 1st quadrant cosθ\cos \theta is maximum at cos0\cos 0{}^\circ which is the minimum angle possible and value of cos0\cos 0{}^\circ is 1. Also, with the increase in angle the value of cosθ\cos \theta decreases.

The value of cos90\cos 90{}^\circ is 0 and the value of cos180\cos 180{}^\circ is -1 which is the minimum value possible for cosθ\cos \theta .
Also, we know that the value of cosθ\cos \theta varies from 1 to -1.

This picture is showing the graph of cosθ\cos \theta .

From graph of cosθ\cos \theta it is clear that the maximum value of cosθ\cos \theta is at 00{}^\circ which is 1 and the minimum value of cosθ\cos \theta is at 180180{}^\circ which is -1.

Also, we know that if we subtract any greater integer number with any smaller integer number we always get the result as a negative number.

Let us assume that a and b are two positive integer numbers and also a < b.
Then when we subtract (a – b) we get a negative number.

Now, if we analyse the value of cos2\cos 2{}^\circ and cos11\cos 11{}^\circ then keeping above discussion in mind we can conclude that the value of cos2\cos 2{}^\circ is greater than the value of cos11\cos 11{}^\circ .

As the angle cos2\cos 2{}^\circ is smaller than cos11\cos 11{}^\circ . So, it is clear that the value of cos2\cos 2{}^\circ is always greater than cos11\cos 11{}^\circ . Also, cos2\cos 2{}^\circ and cos11\cos 11{}^\circ both lie in the first quadrant and all the values in the first quadrant are always positive.

So, if we subtract cos2\cos 2{}^\circ from cos11\cos 11{}^\circ then we are subtracting a smaller value from a greater value. So the resulting value is always a negative number and we know that negative numbers are smaller than 0 as they are lying on the left side of 0 in the number line.

This picture is showing a number line of integers.

So, from all the above discussion we can conclude that
cos11cos2<0\cos 11{}^\circ -\cos 2{}^\circ <0
Or, when we subtract cos2\cos 2{}^\circ from cos11\cos 11{}^\circ we will get a negative value which is always less than 0 in the first quadrant.

But in question, the statement is cos11cos2>0\cos 11{}^\circ -\cos 2{}^\circ >0 which is incorrect.

Thus, the given statement is false.

Note: This is a basic question and if you know the trend of cosθ\cos \theta a different angle then you can even solve this question without writing a single word.

The value of cosθ\cos \theta decreases with increase the value of θ\theta .
cosθ\cos \theta is maximum at cos0\cos 0{}^\circ and cosθ\cos \theta is minimum at cos180\cos 180{}^\circ .
cos0=1\cos 0{}^\circ =1 and cos180=1\cos 180{}^\circ =-1

You may directly calculate the value of cos11\cos 11{}^\circ and cos2\cos 2{}^\circ using a calculator.
cos11cos2<0 0.980.99<0 0.01<0 \begin{aligned} & \Rightarrow \cos 11{}^\circ -\cos 2{}^\circ <0 \\\ & \Rightarrow 0.98-0.99<0 \\\ & \Rightarrow -0.01<0 \\\ \end{aligned}

Thus, the given statement in the question is false.