Finding exact value of a trig function

[Velvet]

Hello!

If the sin u= 3/5 and cos v= (-5)/13 and both angles are in quad. II, can we say that:
Cos u = x/r = (-4)/5 and sin v = y/r= 12/13?

If this is right, then is my reasoning right when I say that:

cos (u + v) = [(-4)/5 x (-5)/13] + 3/5 x 12/13 = 728/845 ?

The number seems pretty big, so I would assume there is an error, but I can't find it...

Also, I have troubles finding X in a tan equation.
If tan x = -1 , can we say that the solutions for x are 135 + 360n and 315 + 360n (all in degrees)?

Thanks for helping!

~V.

 

[soroban]

Hello, Velvet!

If \(\displaystyle \sin u\,=\,\frac{3}{5}\) and \(\displaystyle \cos v\,=\,-\frac{5}{13}\)and both angles are in Quadrant II,
can we say that: .\(\displaystyle \cos u\,=\,\frac{x}{r}\,=\,-\frac{4}{5}\) and \(\displaystyle \sin v\,=\,\frac{y}{r}\,=\,\frac{12}{13}\) ? . . . . yes!

If this is right, then is my reasoning right when I say that:

\(\displaystyle \cos (u\,+\,v)\:=\:-\left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)\,+\,\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)\:=\:\frac{728}{845}\) . . . . This can be reduced

The number seems pretty big, so I would assume there is an error, but I can't find it...

Not very "big" . . . it's still less than 1.

You should have had: .\(\displaystyle \frac{20}{65}\,+\,\frac{36}{65}\:=\:\frac{56}{65}\)

If \(\displaystyle \tan x\,=\,-1\), can we say that the solutions are:
. . . \(\displaystyle x\:=\:135^o\,+\,n\cdot360^o\) and \(\displaystyle 315^o\,+\,n\cdot360^o\) ? . . . . yes!

Even better: .\(\displaystyle x\:=\:135^o\,+\,n\cdot180^o\)

 

[Velvet]

Thank you!

~V.