# Proving trig equation?

• December 13th 2012, 06:21 PM
homeylova223
Proving trig equation?
How would I prove the following.

arcsin(4/5)+arctan(3/4)=pi/2

Any help is appreciated.
• December 13th 2012, 06:36 PM
MarkFL
Re: Proving trig equation?
You could draw a 3-4-5 right triangle and:

a) find that $\arcsin \left( \frac{4}{5} \right)= \text{arccot} \left( \frac{3}{4} \right)$

b) find that $\arctan \left( \frac{3}{4} \right)= \arccos \left( \frac{4}{5} \right)$

Then use the fact that the sum of two complementary inverse trig functions having the same argument is always $\frac{\pi}{2}$.
• December 13th 2012, 06:50 PM
homeylova223
Re: Proving trig equation?
I see why is it that the sum of two complementary inverse trig functions is pi/2?
• December 13th 2012, 06:54 PM
MarkFL
Re: Proving trig equation?
Because by definition, they represent complementary angles.
• December 13th 2012, 07:01 PM
homeylova223
Re: Proving trig equation?
I see thanks.

I have a similar question.

It is

arcsin(3/5)+arcsin(15/17)=(-13/85)

How would I explain this problem. I know sin(a+b)=sinacosb+cosasinb.

But I am not sure what to do.
• December 13th 2012, 07:18 PM
MarkFL
Re: Proving trig equation?
What you mean is:

$\arcsin\left(\frac{3}{5} \right)+\arcsin\left(\frac{15}{17} \right)=\arccos\left(-\frac{13}{85} \right)$

There are two ways we could proceed:

a) Take the sine of both sides:

$\sin\left(\arcsin\left(\frac{3}{5} \right)+\arcsin\left(\frac{15}{17} \right) \right)=\sin\left(\arccos\left(-\frac{13}{85} \right) \right)$

$\sin\left(\arcsin\left(\frac{3}{5} \right) \right)\cos\left(\arcsin\left(\frac{15}{17} \right) \right)+\cos\left(\arcsin\left(\frac{3}{5} \right) \right)\sin\left(\arcsin\left(\frac{15}{17} \right) \right)=\frac{\sqrt{85^2-(-13)^2}}{85}$

$\frac{3}{5}\cdot\frac{\sqrt{17^2-15^2}}{17}+\frac{\sqrt{5^2-3^2}}{5}\cdot\frac{15}{17}=\frac{84}{85}$

$\frac{3}{5}\cdot\frac{8}{17}+\frac{4}{5}\cdot\frac {15}{17}=\frac{84}{85}$

$\frac{24+60}{85}=\frac{84}{85}$

$\frac{84}{85}=\frac{84}{85}$

b) Take the cosine of both sides: