Question

If $\cos x=\tan y,\cos y=\tan z,\cos z=\tan x$ , then prove that $\sin x=\sin y=\sin z=2\sin 18{}^\circ $ .

Answer 1

Hint: First, we have to take any one relation to start, so we will take $\cos x=\tan y$ . Now, we have to convert tan function in such a way that we get all the equations and terms in form of sin function only. So, here we will be using some identities like \[\left( 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \right)\] , \[\sec \theta =\dfrac{1}{\cos \theta }\] , \[\cot \theta =\dfrac{1}{\tan \theta }\] , \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] . Using all these identities, we will get a quadratic equation whose roots we can find using the formula \[D=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . At last we will compare our root with the value of \[\sin 18{}^\circ =\left( \dfrac{\sqrt{5}-1}{4} \right)\] . Thus, we will be able to prove the relation given in question.

Complete step-by-step answer:
Here, first we will take the relation as $\cos x=\tan y$ .
On squaring, both sides of equation, we get as
${{\cos }^{2}}x={{\tan }^{2}}y$
Now, we know that \[\left( 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \right)\] . So, from this substituting value of tan we get,
\[\Rightarrow {{\cos }^{2}}x={{\sec }^{2}}y-1\]
Also, we know that \[\sec \theta =\dfrac{1}{\cos \theta }\] so, using this relation and on rearranging the terms, we can write equation as
\[\Rightarrow {{\cos }^{2}}x+1=\dfrac{1}{{{\cos }^{2}}y}\]
Now, in the question, it is given that \[\cos y=\tan z\] so, we will put \[\tan z\] in place of \[\cos y\] . So, we get as
\[\Rightarrow {{\cos }^{2}}x+1=\dfrac{1}{{{\tan }^{2}}z}\]
Now, we will use the relation \[\cot \theta =\dfrac{1}{\tan \theta }\] so, we will substitute this value in the above equation, we get as
\[\Rightarrow {{\cos }^{2}}x+1={{\cot }^{2}}z\]
We know that \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] so, using this we get the equation as
\[\Rightarrow 1-{{\sin }^{2}}x+1=\dfrac{{{\cos }^{2}}z}{{{\sin }^{2}}z}\]
Again we will be using the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and placing value of \[{{\cos }^{2}}\theta \] in the equation, we get
\[\Rightarrow 2-{{\sin }^{2}}x=\dfrac{{{\cos }^{2}}z}{1-{{\cos }^{2}}z}\]
Now, we are given that \[\cos z=\tan x\] .So, putting this value we get
\[\Rightarrow 2-{{\sin }^{2}}x=\dfrac{{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\]
We will substitute the value of tan as \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] . we will get as
\[\Rightarrow 2-{{\sin }^{2}}x=\dfrac{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}\]
On further simplifying and cancelling the denominator, we get
\[\Rightarrow 2-{{\sin }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}\]
\[\Rightarrow \left( 2-{{\sin }^{2}}x \right)\left( 1-2{{\sin }^{2}}x \right)={{\sin }^{2}}x\]
 (replacing \[{{\cos }^{2}}x=1-{{\sin }^{2}}x\] and taking denominator to LHS)
Now, we will assume \[{{\sin }^{2}}x=t\] and on substituting we get equation as
\[\Rightarrow \left( 2-t \right)\left( 1-2t \right)=t\]
\[\Rightarrow 2-4t-t+2{{t}^{2}}=t\]
\[\Rightarrow 2{{t}^{2}}-6t+2=0\]
This we got quadratic equation and roots we can find by using the formula \[D=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
 So, on using this i.e. \[a=2,b=-6,c=2\] we get roots as
\[\Rightarrow t=\dfrac{6\pm \sqrt{{{6}^{2}}-4\cdot 2\cdot 2}}{2\cdot 2}\]
On further solving, we get
\[\Rightarrow t=\dfrac{6\pm \sqrt{36-16}}{4}\]
\[\Rightarrow t=\dfrac{6\pm \sqrt{20}}{4}\]
\[\Rightarrow t=\dfrac{6\pm 2\sqrt{5}}{4}\]
\[\Rightarrow t=\dfrac{6+2\sqrt{5}}{4},\dfrac{6-2\sqrt{5}}{4}\]
No, we know that the range of sin function is \[\left[ -1,1 \right]\] so, root \[\dfrac{6+2\sqrt{5}}{4}>1\] . So, this will not be root
\[\therefore t=\dfrac{6-2\sqrt{5}}{4}\]
Replacing value of t and we should know that \[{{\left( \dfrac{\sqrt{5}-1}{2} \right)}^{2}}=\dfrac{6-2\sqrt{5}}{4}\] . So, putting the values, we get
\[\therefore {{\sin }^{2}}x={{\left( \dfrac{\sqrt{5}-1}{2} \right)}^{2}}\]
\[\therefore \sin x=\left( \dfrac{\sqrt{5}-1}{2} \right)\]
Now, we know that \[\sin 18{}^\circ =\left( \dfrac{\sqrt{5}-1}{4} \right)\] which can be written as \[\sin 18{}^\circ =\left( \dfrac{\sqrt{5}-1}{2} \right)\times \dfrac{1}{2}\] .So, we get as \[2\sin 18{}^\circ =\left( \dfrac{\sqrt{5}-1}{2} \right)\]
Thus, we can write as \[\sin x=2\sin 18{}^\circ \] .
Similarly, we can prove \[\sin y=2\sin 18{}^\circ ,\sin z=2\sin 18{}^\circ \] .
Hence, proved.

Note: Be careful while converting all the trigonometric functions like cos, tan, cot, sec into sine functions. One should know all the identities used here otherwise if one sign i.e. negative or positive sign is interchanged the whole answer will be wrong and will not be able to prove the relation given in question. Also, the value of \[\sin 18{}^\circ =\left( \dfrac{\sqrt{5}-1}{4} \right)\] should be known. Rest all part is simple solving quadratic equations.