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

Here we are going to solve the sum and find the value of$\alpha$, we need trigonometric identities to find the value.

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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 $\tan \dfrac{\pi }{8} = \sqrt 2 - 1$