Application of a Sum Formula

**Bashyboy** · January 6th 2012, 06:03 AM

"Write cos(arctan 1 + arccos x) as an algebraic expression." So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and
v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as
cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as
1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did.

**e^(i*pi)** · January 6th 2012, 06:13 AM

Originally Posted by Bashyboy

"Write cos(arctan 1 + arccos x) as an algebraic expression." So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and
v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as
cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as
1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did.

pi/4 is on the unit circle and its cosine is not 0: $\cos \left(\frac{\pi}{4}\right) = \dfrac{\sqrt2}{2}$ .

**Plato** · January 6th 2012, 06:14 AM

Originally Posted by Bashyboy

"Write cos(arctan 1 + arccos x) as an algebraic expression." So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and
v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as 1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did.

This is standard fair: $\arctan (1) = \frac{\pi }{4}\;\& \,\cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$

**Bashyboy** · January 6th 2012, 06:43 AM

You two are certainly right--and I am grateful for that. Sorry for the imprudent observation on my part.

**Prove It** · January 6th 2012, 05:01 PM

Originally Posted by Bashyboy

"Write cos(arctan 1 + arccos x) as an algebraic expression." So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and
v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as
cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as
1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did.

An alternative is to note that

$\displaystyle \begin{align*} \sin^2{X} + \cos^2{X} &\equiv 1 \\ \frac{\sin^2{X}}{\cos^2{X}} + \frac{\cos^2{X}}{\cos^2{X}} &\equiv \frac{1}{\cos^2{X}} \\ \tan^2{X} + 1 &\equiv \frac{1}{\cos^2{X}} \\ \frac{1}{\tan^2{X} + 1} &\equiv \cos^2{X} \\ \cos{X} &\equiv \sqrt{\frac{1}{\tan^2{X} + 1}} \end{align*}$

So that means

$\displaystyle \begin{align*} \cos{\left(\arctan{1}\right)} &= \sqrt{\frac{1}{\left[\tan{\left(\arctan{1}\right)}\right]^2 + 1}} \\ &= \sqrt{\frac{1}{1^2 + 1}} \\ &= \sqrt{\frac{1}{1 + 1}} \\ &= \sqrt{\frac{1}{2}} \\ &= \frac{\sqrt{1}}{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} \end{align*}$

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