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Question: If \( \sin 32{}^\circ =k \) and \( \cos x=1-2{{k}^{2}} \) ; \( \alpha ,\beta \) are the values of x ...

If sin⁡32∘=k\sin 32{}^\circ =k and cos⁡x=1−2k2\cos x=1-2{{k}^{2}} ; α,β\alpha ,\beta are the values of x between 0∘0{}^\circ and 360∘360{}^\circ with α<β\alpha <\beta , then the value in degrees of βα=37k\dfrac{\beta }{\alpha }=\dfrac{37}{k} . Find the value of k.

Explanation

Solution

Hint : Substitute sin⁡32∘=k\sin 32{}^\circ =k in cos⁡x=1−2k2\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. 1−2sin⁡2A=cos⁡2A1-2{{\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∘=k\sin 32{}^\circ =k and cos⁡x=1−2k2\cos x=1-2{{k}^{2}} ; α,β\alpha ,\beta are the values of x between 0∘0{}^\circ and 360∘360{}^\circ with α<β\alpha <\beta , then the value in degrees of βα=37k\dfrac{\beta }{\alpha }=\dfrac{37}{k} .
We need to find the value of k.
Given sin⁡32∘=k\sin 32{}^\circ =k …(1)
And, cos⁡x=1−2k2\cos x=1-2{{k}^{2}} …(2)
Substituting equation (1) in equation (2), we will get the following:
cos⁡x=1−2sin⁡232∘\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. 1−2sin⁡2A=cos⁡2A1-2{{\sin }^{2}}A=\cos 2A
Using this property on the above equation, we will get the following:
cos⁡x=1−2sin⁡232∘\cos x=1-2{{\sin }^{2}}32{}^\circ
cos⁡x=cos⁡64∘\cos x=\cos 64{}^\circ
Since, we know that the value of x lies between 0∘0{}^\circ and 360∘360{}^\circ ,
Hence, x will be equal to the following values:
x=64∘,296∘x=64{}^\circ ,296{}^\circ
Now, since we are given that α<β\alpha <\beta
So, α=64∘\alpha =64{}^\circ and β=296∘\beta =296{}^\circ
Now, we will find the value of βα\dfrac{\beta }{\alpha }
Hence, the value of k is 8∘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. 1−2sin⁡2A=cos⁡2A1-2{{\sin }^{2}}A=\cos 2A . Students often confuse and write the difference as twice the cosine of angle 2A, i.e. 1−2sin⁡2A=2cos⁡2A1-2{{\sin }^{2}}A=2\cos 2A , which is wrong. Such mistakes should be avoided, as they lead to wrong answers.