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Question: The value of \(\cos {{9}^{0}}-\sin {{9}^{0}}\) is: (a) \(-\dfrac{\sqrt{5-\sqrt{5}}}{2}\) (b) \(...

The value of cos⁡90−sin⁡90\cos {{9}^{0}}-\sin {{9}^{0}} is:
(a) −5−52-\dfrac{\sqrt{5-\sqrt{5}}}{2}
(b) 5+54\dfrac{5+\sqrt{5}}{4}
(c) 125−5\dfrac{1}{2}\sqrt{5-\sqrt{5}}
(d) None of these

Explanation

Solution

Hint:First of all multiply and divide the given expression by 2\sqrt{2} then the expression will look like 2(12cos⁡90−12sin⁡90)\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos {{9}^{0}}-\dfrac{1}{\sqrt{2}}\sin {{9}^{0}} \right).Now, in this expression we can write 12\dfrac{1}{\sqrt{2}} as cos⁡450\cos {{45}^{0}} or sin⁡450\sin {{45}^{0}} so we can write the expression as 2(sin⁡450cos⁡90−cos⁡450sin⁡90)\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right) which can be further written as 2(sin⁡(450−90))\sqrt{2}\left( \sin \left( {{45}^{0}}-{{9}^{0}} \right) \right) which is equal to 2sin⁡360\sqrt{2}\sin {{36}^{0}} then put the value of sin⁡360\sin {{36}^{0}}.

Complete step-by-step answer:
The expression that we have to evaluate is:
cos⁡90−sin⁡90\cos {{9}^{0}}-\sin {{9}^{0}}
Multiplying and dividing the above expression by 2\sqrt{2} will give us the following expression.
2(12cos⁡90−12sin⁡90)\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos {{9}^{0}}-\dfrac{1}{\sqrt{2}}\sin {{9}^{0}} \right)
We know from the value of trigonometric ratios that the value of sin⁡450&cos⁡450\sin {{45}^{0}}\And \cos {{45}^{0}} is equal to 12\dfrac{1}{\sqrt{2}} so substituting the value of 12\dfrac{1}{\sqrt{2}} in the above expression we get,
2(sin⁡450cos⁡90−cos⁡450sin⁡90)\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right) …………Eq. (1)
If you can see carefully the above expression, you recall that the above expression apart from 2\sqrt{2} is the expansion of sin⁡(450−90)\sin \left( {{45}^{0}}-{{9}^{0}} \right) which is equal to sin⁡450cos⁡90−cos⁡450sin⁡90\sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} so writing this expression in as sin⁡(450−90)\sin \left( {{45}^{0}}-{{9}^{0}} \right) in the above we get,
2sin⁡(450−90)\sqrt{2}\sin \left( {{45}^{0}}-{{9}^{0}} \right)
Simplifying the above expression we get,
2sin⁡360\sqrt{2}\sin {{36}^{0}}
We know from the trigonometric ratios that the value of sin⁡360\sin {{36}^{0}} is equal to 10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4} so substituting this value of sin⁡360\sin {{36}^{0}} in the above expression we get,
2(10−254)\sqrt{2}\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right)
Now, taking 2\sqrt{2} as common from the expression in the under root we get,
2(2(5)−254) =2(2)(5−54) =2(5−54) =5−52 \begin{aligned} & \sqrt{2}\left( \dfrac{\sqrt{2\left( 5 \right)-2\sqrt{5}}}{4} \right) \\\ & =\sqrt{2}\left( \sqrt{2} \right)\left( \dfrac{\sqrt{5-\sqrt{5}}}{4} \right) \\\ & =2\left( \dfrac{\sqrt{5-\sqrt{5}}}{4} \right) \\\ & =\dfrac{\sqrt{5-\sqrt{5}}}{2} \\\ \end{aligned}
From the above solution, the evaluation of the given expression is 5−52\dfrac{\sqrt{5-\sqrt{5}}}{2}.
Hence, the correct option is (c).

Note: The alternate way of solving the above problem is to rewrite the eq. (1) in the above solution as:
2(sin⁡450cos⁡90−cos⁡450sin⁡90)\sqrt{2}\left( \sin {{45}^{0}}\cos {{9}^{0}}-\cos {{45}^{0}}\sin {{9}^{0}} \right)
In place of this expression we can write the above expression as:
2(cos⁡450cos⁡90−sin⁡450sin⁡90)\sqrt{2}\left( \cos {{45}^{0}}\cos {{9}^{0}}-\sin {{45}^{0}}\sin {{9}^{0}} \right)
Now, the above trigonometric expression i.e. cos⁡450cos⁡90−sin⁡450sin⁡90\cos {{45}^{0}}\cos {{9}^{0}}-\sin {{45}^{0}}\sin {{9}^{0}} is the expansion of cos⁡(450+90)\cos \left( {{45}^{0}}+{{9}^{0}} \right) so we can substitute this value of expansion in the above expression.
2cos⁡(450+90)\sqrt{2}\cos \left( {{45}^{0}}+{{9}^{0}} \right)
Simplifying the above expression we get,
2cos⁡540\sqrt{2}\cos {{54}^{0}}
We know from the trigonometric ratios that the value of cos⁡540\cos {{54}^{0}} is equal to 10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4} we get,
2(10−254) =5−52 \begin{aligned} & \sqrt{2}\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right) \\\ & =\dfrac{\sqrt{5-\sqrt{5}}}{2} \\\ \end{aligned}
Hence, we have got the same value of the expression in the given question as we have solved above.We should remember the formula of sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B)=\sin A \cos B-\cos A \sin B and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A \cos B-\sin A \sin B.Also we have to memorize the trigonometric angle value of sin⁡360\sin {{36}^{0}} is 10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4} and cos⁡540\cos {{54}^{0}} is 10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4}