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Question: The value of \[\sin {{27}^{0}}-\cos {{27}^{0}}\]is equal to A) \[-\dfrac{\sqrt{3-\sqrt{5}}}{2}\] ...

The value of sin⁡270−cos⁡270\sin {{27}^{0}}-\cos {{27}^{0}}is equal to
A) −3−52-\dfrac{\sqrt{3-\sqrt{5}}}{2}
B) −5−52-\dfrac{\sqrt{5-\sqrt{5}}}{2}
C) −5−122-\dfrac{\sqrt{5}-1}{2\sqrt{2}}
D) 3−52\dfrac{\sqrt{3-\sqrt{5}}}{2}

Explanation

Solution

In the given question, we have been asked to find the value of the given expression. In order to find the value, first we need to rewrite the equation in the sine form then by applying the trigonometric identity of sum or differences of product of sine function i.e. sin⁡a−sin⁡b=2cos⁡(a+b2)sin⁡(a−b2)\sin a-\sin b=2\cos \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right) and we will simplify the given expression. Then using the trigonometric ratios table we will put the value and solve the expression. In this way we will get the required result.

Complete step-by-step solution:
We have given that,
⇒sin⁡270−cos⁡270\Rightarrow \sin {{27}^{0}}-\cos {{27}^{0}}
Rewritten the above expression as,
⇒sin⁡270−cos⁡(90−630)\Rightarrow \sin {{27}^{0}}-\cos \left( 90-{{63}^{0}} \right)
Using the property of trigonometry that is if one angle is 90 degrees we can easily figure out;
sin⁡θ=cos⁡(90−θ)\sin \theta =\cos \left( 90-\theta \right)And cos⁡θ=sin⁡(90−θ)\cos \theta =\sin \left( 90-\theta \right)
Applying the identity in the above expression, we get
⇒sin⁡270−sin⁡630\Rightarrow \sin {{27}^{0}}-\sin {{63}^{0}}
Using the trigonometry identity of difference or sum as product i.e.
sin⁡a−sin⁡b=2cos⁡(a+b2)sin⁡(a−b2)\sin a-\sin b=2\cos \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)
Applying the identity in the above expression, we get
⇒sin⁡270−sin⁡630=2cos⁡(27+632)sin⁡(27−632)\Rightarrow \sin {{27}^{0}}-\sin {{63}^{0}}=2\cos \left( \dfrac{27+63}{2} \right)\sin \left( \dfrac{27-63}{2} \right)
Solving the above expression, we get
⇒2cos⁡450sin⁡(−360)=−2cos⁡450sin⁡360\Rightarrow 2\cos {{45}^{0}}\sin \left( -{{36}^{0}} \right)=-2\cos {{45}^{0}}\sin {{36}^{0}}
Using the trigonometric ratios table;
The value of cos⁡450=12\cos {{45}^{0}}=\dfrac{1}{\sqrt{2}}.
Putting the value, we get
⇒−2×12×sin⁡360\Rightarrow -2\times \dfrac{1}{\sqrt{2}}\times \sin {{36}^{0}}
Now solving,
⇒sin⁡360\Rightarrow \sin {{36}^{0}}
⇒sin⁡360=1−cos⁡2(36)0\Rightarrow \sin {{36}^{0}}=\sqrt{1-{{\cos }^{2}}{{\left( 36 \right)}^{0}}}
As,
Now solving,
cos⁡360\cos {{36}^{0}}
⇒cos⁡360=cos⁡2(180)\Rightarrow \cos {{36}^{0}}=\cos 2\left( {{18}^{0}} \right)
Using the identity, cos⁡2x=1−2sin⁡2x\cos 2x=1-2{{\sin }^{2}}x
⇒cos⁡2(180)=1−2(sin⁡2(18)0)\Rightarrow \cos 2\left( {{18}^{0}} \right)=1-2\left( {{\sin }^{2}}{{\left( 18 \right)}^{0}} \right)
Put the value of sin⁡2(18)0=5+14{{\sin }^{2}}{{\left( 18 \right)}^{0}}=\dfrac{\sqrt{5}+1}{4}
⇒1−2(5+14)2\Rightarrow 1-2{{\left( \dfrac{\sqrt{5}+1}{4} \right)}^{2}}
Using the identity, (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab
\Rightarrow 1-2\left( \dfrac{5+1-2\sqrt{5}}{16} \right)$$$$\Rightarrow 1-2\left( \dfrac{6-2\sqrt{5}}{16} \right)
⇒cos⁡360=5+14\Rightarrow \cos {{36}^{0}}=\dfrac{\sqrt{5}+1}{4}
Putting the value of cos⁡360=5+14\cos {{36}^{0}}=\dfrac{\sqrt{5}+1}{4},
\Rightarrow \sin {{36}^{0}}=\sqrt{1-{{\left( \dfrac{\sqrt{5}+1}{4} \right)}^{2}}}$$$$\Rightarrow \sin {{36}^{0}}=\sqrt{1-\left( \dfrac{5+1+2\sqrt{5}}{16} \right)}$$$$\Rightarrow \sin {{36}^{0}}=\dfrac{1}{4}\times \sqrt{10-2\sqrt{5}}
∴sin⁡360=10−254\therefore \sin {{36}^{0}}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}
Now, putting the value of sin⁡360=10−254\sin {{36}^{0}}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}in plug-in solved expression i.e.
⇒−2×12×sin⁡360\Rightarrow -2\times \dfrac{1}{\sqrt{2}}\times \sin {{36}^{0}}
⇒−2×12×10−254\Rightarrow -2\times \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{10-2\sqrt{5}}}{4}
Simplifying the above expression, we get
⇒−2×12×10−252×2×2\Rightarrow -\sqrt{2}\times \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{10-2\sqrt{5}}}{2\times \sqrt{2}\times \sqrt{2}}
Cancelling out the common terms, we get
⇒−2×12×5−52×2\Rightarrow -\sqrt{2}\times \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{5-\sqrt{5}}}{2\times \sqrt{2}}
⇒−5−52\Rightarrow -\dfrac{\sqrt{5-\sqrt{5}}}{2}
Thus,
⇒sin⁡270−cos⁡270=−5−52\Rightarrow \sin {{27}^{0}}-\cos {{27}^{0}}=-\dfrac{\sqrt{5-\sqrt{5}}}{2}

Hence, the option (B) is the correct answer.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.