Question

If \[x = {\log _{0.75}}(\dfrac{{148}}{{111}})\] and \[y = \dfrac{{\cos (\dfrac{\pi }{4} - \theta ) - \cos (\dfrac{\pi }{4} + \theta )}}{{\sin (\dfrac{{2\pi }}{3} + \theta ) - \sin (\dfrac{{2\pi }}{3} - \theta )}}\] and \[x - y = \tan \alpha \], Then find $\alpha.$

Answer 1

Hint: Here we are going to solve the sum and find the value of\[\alpha \], we need trigonometric identities to find the value.

Formula used: 

\[\log \dfrac{a}{b} = - \log \dfrac{b}{a}\]
\[\sin (A + B) = \sin A\cos B + \cos A\sin B\]
\[\cos (A + B) = \cos A\cos B - \sin A\sin B\]
\[\sin (A - B) = \sin A\cos B - \cos A\sin B\]
\[\cos (A - B) = \cos A\cos B + \sin A\sin B\]

Complete step by step answer:
It is given that, \[x = {\log _{0.75}}(\dfrac{{148}}{{111}})\] and \[y = \dfrac{{\cos (\dfrac{\pi }{4} - \theta ) - \cos (\dfrac{\pi }{4} + \theta )}}{{\sin (\dfrac{{2\pi }}{3} + \theta ) - \sin (\dfrac{{2\pi }}{3} - \theta )}}\] and \[x - y = \tan \alpha \],
First, we consider the term to solve it for a simple value,
\[x = {\log _{0.75}}(\dfrac{{148}}{{111}})\]
We have to simplify the above equation then we get,
\[x = {\log _{\dfrac{{75}}{{100}}}}(\dfrac{{37 \times 4}}{{37 \times 3}})\]
Further we have to eliminate same term from the numerator and denominator so that we can get the following equation,
 \[x = {\log _{\dfrac{3}{4}}}(\dfrac{4}{3})\]
By the property of logarithm\[\log \dfrac{a}{b} = - \log \dfrac{b}{a}\] ,
\[x = - {\log _{\dfrac{3}{4}}}(\dfrac{3}{4})\]
We also know that if the base and the power of a logarithm is same, its value will be one, which is nothing but the following property \[{\log _a}a = 1\],
\[x = - 1\]
Now, we consider the term y and solve it further,
\[y = \dfrac{{\cos (\dfrac{\pi }{4} - \theta ) - \cos (\dfrac{\pi }{4} + \theta )}}{{\sin (\dfrac{{2\pi }}{3} + \theta ) - \sin (\dfrac{{2\pi }}{3} - \theta )}}\]
Applying the trigonometric identities we have the equation rewritten as follows,
\[y = \dfrac{{[\cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta ] - [\cos \dfrac{\pi }{4}\cos \theta - \sin \dfrac{\pi }{4}\sin \theta ]}}{{[\sin \dfrac{{2\pi }}{3}\cos \theta + \cos \dfrac{{2\pi }}{3}\sin \theta ] - [\sin \dfrac{{2\pi }}{3}\cos \theta - \cos \dfrac{{2\pi }}{3}\sin \theta ]}}\]
Now let us simplify the addition and subtraction in the above equation, we get,
\[y = \dfrac{{\cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta - \cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta }}{{\sin \dfrac{{2\pi }}{3}\cos \theta + \cos \dfrac{{2\pi }}{3}\sin \theta - \sin \dfrac{{2\pi }}{3}\cos \theta + \cos \dfrac{{2\pi }}{3}\sin \theta }}\]
On further solving the above equation we get,
\[y = \dfrac{{2\sin \dfrac{\pi }{4}\sin \theta }}{{2\cos \dfrac{{2\pi }}{3}\sin \theta }}\]
Let us eliminate the similar terms from the numerator and denominator we get,
\[y = \dfrac{{\sin \dfrac{\pi }{4}\sin \theta }}{{\cos \dfrac{{2\pi }}{3}\sin \theta }}\]
Now let us substitute the values of \[\sin \dfrac{\pi }{4} = \sqrt 2 ,\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2},\]in the above equation, we get,
\[y = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{ - 1}}{2}}}\]
Let us simplify the equation which we derived above, we get the value of y as,
\[y = - \sqrt 2 \]
From the given condition we know that
\[x - y = \tan \alpha \]
Let us now substitute the values of \[x = - 1\] and \[y = - \sqrt 2 \] in the above equation, we get,
\[\tan \alpha = \sqrt 2 - 1\]
We know that the value of \[\tan \dfrac{\pi }{8} = \sqrt 2 - 1\]
So, \[\alpha = \dfrac{\pi }{8}\]
Hence,

Therefore, The value of \[\alpha = \dfrac{\pi }{8}\].

Note:
Here is the calculation for \[\tan \dfrac{\pi }{8}\]
We know that, \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]
Let us take, \[\tan \dfrac{\pi }{8} = x\]
Now, \[\tan \dfrac{\pi }{4} = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
Substitute the value in the above equation we get,
\[1 = \dfrac{{2x}}{{1 - {x^2}}}\]
On solving the above equation we get,
\[{x^2} + 2x - 1 = 0\]
(Sreedhar Acharya’s formula: for a quadratic equation \[a{x^2} + bx + c = 0\] the root is found using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\])
 Now applying the formula mentioned above we get,
\[x = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2}\]
We will consider the positive value,
\[x = \dfrac{{ - 2 + \sqrt 8 }}{2} = \sqrt 2 - 1\]
So, the value of \[\tan \dfrac{\pi }{8} = \sqrt 2 - 1\]