# If a line makes an angle $\alpha $,$\beta $,$\gamma $ with the coordinate axes. Prove that $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 = 0$.

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Hint: Here we use direction cosines of a line making angles with coordinate axes and its property is also used to solve the problem. We will let the direction cosines and use them.

Now $\alpha ,\beta ,\gamma $ are the angle which the line makes with the co-ordinate axis. So, the direction cosines of the line are

Direction – cosines = $\cos \alpha ,\cos \beta ,\cos \gamma $

Now, as we know the direction cosines of a line are \[l,m,n\]. So, we can write direction cosines as,

$l = \cos \alpha $, $m = \cos \beta $, $n = \cos \gamma $ ……. (1)

Now, using the property of direction-cosines which is ${l^2} + {m^2} + {n^2} = 1$. Putting the values of $l,m,n$ from equation (1) in the property.

Putting $l = \cos \alpha $, $m = \cos \beta $, $n = \cos \gamma $, we get

$ \Rightarrow $${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$ ………. (2)

Now, taking the L. H. S term of the question,

L. H. S = $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1$ ……. (3)

From trigonometric identities, we know that $\cos 2x = 2{\cos ^2}x - 1$, applying this property in equation (3), we get

L. H. S = $(2{\cos ^2}\alpha - 1) + (2{\cos ^2}\beta - 1) + (2{\cos ^2}\gamma - 1) + 1$

Simplifying the above term,

L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 3 + 1$

L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 2$ ……… (4)

Now, from equation (2) putting the value of ${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $ in equation (4), we get

L. H. S = $2(1) - 2 = 0$ = R. H. S

Hence, Proved.

Note: Don’t confuse between the direction ratios and direction cosines. They both look similar but actually they are different. Direction ratios are obtained when we divide the direction cosines by their magnitude. Also, it is recommended to learn trigonometric identities of $\cos 2x$, $\sin 2x$, $\cos 3x$, $\sin 3x$ which are helpful in solving these types of questions. Proper use of identities led to proper solutions in less time without any mistakes.

Now $\alpha ,\beta ,\gamma $ are the angle which the line makes with the co-ordinate axis. So, the direction cosines of the line are

Direction – cosines = $\cos \alpha ,\cos \beta ,\cos \gamma $

Now, as we know the direction cosines of a line are \[l,m,n\]. So, we can write direction cosines as,

$l = \cos \alpha $, $m = \cos \beta $, $n = \cos \gamma $ ……. (1)

Now, using the property of direction-cosines which is ${l^2} + {m^2} + {n^2} = 1$. Putting the values of $l,m,n$ from equation (1) in the property.

Putting $l = \cos \alpha $, $m = \cos \beta $, $n = \cos \gamma $, we get

$ \Rightarrow $${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$ ………. (2)

Now, taking the L. H. S term of the question,

L. H. S = $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1$ ……. (3)

From trigonometric identities, we know that $\cos 2x = 2{\cos ^2}x - 1$, applying this property in equation (3), we get

L. H. S = $(2{\cos ^2}\alpha - 1) + (2{\cos ^2}\beta - 1) + (2{\cos ^2}\gamma - 1) + 1$

Simplifying the above term,

L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 3 + 1$

L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 2$ ……… (4)

Now, from equation (2) putting the value of ${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $ in equation (4), we get

L. H. S = $2(1) - 2 = 0$ = R. H. S

Hence, Proved.

Note: Don’t confuse between the direction ratios and direction cosines. They both look similar but actually they are different. Direction ratios are obtained when we divide the direction cosines by their magnitude. Also, it is recommended to learn trigonometric identities of $\cos 2x$, $\sin 2x$, $\cos 3x$, $\sin 3x$ which are helpful in solving these types of questions. Proper use of identities led to proper solutions in less time without any mistakes.

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