# How do you solve #tan^ (-1) (1)+ cos ^(-1)(-1/2) + sin ^-1(-1/2)#?

##### 2 Answers

#### Explanation:

Note that typical trigonometric functions are in the form

Inverse trigonometric functions are then in the form

Then, for the three inverse trig functions we have here, note that:

#alpha=tan^-1(1)" "=>" "tan(alpha)=-1# #beta=cos^-1(-1/2)" "=>" "cos(beta)=-1/2# #phi=sin^-1(-1/2)" "=>" "sin(phi)=-1/2#

where

Our knowledge of trigonometric functions tells us that:

#tan(pi/4)=1# #cos((2pi)/3)=-1/2# #sin(-pi/6)=-1/2#

Note that, for example,

- range of
#alpha=tan^-1(x)# :#" "-pi/2lealphalepi/2# - range of
#beta=cos^-1(x)# :#" "color(white)-0lebetalepi# - range of
#phi=sin^-1(x)# :#" "-pi/2lephilepi/2#

Then:

#alpha=pi/4# #beta=(2pi)/3# #phi=-pi/6#

Thus:

#tan^-1(1)+cos^-1(-1/2)+sin^-1(-1/2)#

#=alpha+beta+phi#

#=pi/4+(2pi)/3-pi/6#

#=(3pi)/4#

Undefined question.

#### Explanation:

We can't solve this question. The question just ask us to add up

3 specific quantities. To me, this problem is trigonometrical undefined. Reasons:

arctan (1) -->

With the same variable arc x, there are no indications about different

domains for periodic functions.

However, we can use calculator to add up the results by ignoring the trig aspect of the question.

45^@ + 120^@ - 30^@ = 135^@