Question

If $ \sin 32{}^\circ =k $ and $ \cos x=1-2{{k}^{2}} $ ; $ \alpha ,\beta $ are the values of x between $ 0{}^\circ $ and $ 360{}^\circ $ with $ \alpha <\beta $ , then the value in degrees of $ \dfrac{\beta }{\alpha }=\dfrac{37}{k} $ . Find the value of k.

Answer 1

Hint: Substitute $ \sin 32{}^\circ =k $ in $ \cos x=1-2{{k}^{2}} $ . Then use the property that for an angle A, the difference of 1 and the double of the square of sine of angle A is equal to cosine of angle 2A. i.e. $ 1-2{{\sin }^{2}}A=\cos 2A $ . Then find the two values of x. Divide them to find the final answer.

Complete step-by-step answer:
In this question, we are given that $ \sin 32{}^\circ =k $ and $ \cos x=1-2{{k}^{2}} $ ; $ \alpha ,\beta $ are the values of x between $ 0{}^\circ $ and $ 360{}^\circ $ with $ \alpha <\beta $ , then the value in degrees of $ \dfrac{\beta }{\alpha }=\dfrac{37}{k} $ .
We need to find the value of k.
Given $ \sin 32{}^\circ =k $ …(1)
And, $ \cos x=1-2{{k}^{2}} $ …(2)
Substituting equation (1) in equation (2), we will get the following:
 $ \cos x=1-2{{\sin }^{2}}32{}^\circ $
We already know that for an angle A, the difference of 1 and the double of the square of sine of angle A is equal to cosine of angle 2A.
i.e. $ 1-2{{\sin }^{2}}A=\cos 2A $
Using this property on the above equation, we will get the following:
 $ \cos x=1-2{{\sin }^{2}}32{}^\circ $
 $ \cos x=\cos 64{}^\circ $
Since, we know that the value of x lies between $ 0{}^\circ $ and $ 360{}^\circ $ ,
Hence, x will be equal to the following values:
 $ x=64{}^\circ ,296{}^\circ $
Now, since we are given that $ \alpha <\beta $
So, $ \alpha =64{}^\circ $ and $ \beta =296{}^\circ $
Now, we will find the value of $ \dfrac{\beta }{\alpha } $
Hence, the value of k is $ 8{}^\circ $
This is the final answer.

Note: In this question, it is very important to know that for an angle A, the difference of 1 and the double of the square of sine of angle A is equal to cosine of angle 2A.
i.e. $ 1-2{{\sin }^{2}}A=\cos 2A $ . Students often confuse and write the difference as twice the cosine of angle 2A, i.e. $ 1-2{{\sin }^{2}}A=2\cos 2A $ , which is wrong. Such mistakes should be avoided, as they lead to wrong answers.