# Thread: Trigonometrical equation

1. ## Trigonometrical equation

Hi. I have a problem solving this trigonometrical equation:
6sin(pi+x)+cos(pi/2+x) if cosx=0.6 and 0<x<pi/2
If you can, please ask questions so that I try to answer them myself and solve the equation step by step. I'm not from an English speaking country, but I hope we have similar terms and you understand me.
Do I reduce first? I get -7sinx, but I'm not sure how to use then cosx=0.6, if I reduce -7sinx back to cos I get -7cos(3pi/2+x) here also I don't know how to use cosx=0.6....

2. ## Re: Trigonometrical equation

If $\cos(x)=0,6$ then $x=\arccos(0,6)$ (1)
Offcourse there're infinite solutions for $x$ responding to that, but there's given that $x\in \]0,\frac{\pi}{2}\[$
Substitute (1) in $6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$

But you can't call it an equation because $6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$ is not equated to anything ... ?

3. ## Re: Trigonometrical equation

Hm, if it's not an equation is it an expression then?
Originally Posted by Siron
If $\cos(x)=0,6$ then $x=\arccos(0,6)$ (1)
Wait, if $\cos(x)=0,6$ then shouldn\t $x=\pmarccos0,6+2\pi n, n\in Z$? What do I do with it?
Oh, so yeah, since 0<x<pi/2, then x=arccos0.6, but still, arccos0.6 doesn't have an exact value...

4. ## Re: Trigonometrical equation

It's indeed not practical to work with, but arccos(0,6)=53,13° or -53,13° because cos(x)=cos(-x)
Is it the intention to calculate it without a calculator? ...

5. ## Re: Trigonometrical equation

I need an exact value , 53.13° is rounded up.
So an exact value would be -7sin(arccos0.6)?

6. ## Re: Trigonometrical equation

Originally Posted by Evaldas
I need an exact value , 53.13° is rounded up.
So an exact value would be -7sin(arccos0.6)?
Almost, $6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)=$ $5\sin(x)$.
So the exact value is $5\sin(\arccos(0,6))$

But this can be simpliefd by usin the fact that $\sin[\arccos(x)]=\sqrt{1-x^2}$

7. ## Re: Trigonometrical equation

Originally Posted by Siron
Almost, $6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)=$ $5\sin(x)$.
So the exact value is $5\sin(\arccos(0,6))$
I got -7sinx because I reduced 6sin(pi+x) and because it's the 3rd quarter (pi<x<3pi/2) sin is negative. I then reduced cos(pi/2+x), since it's 90 degrees it changes its name into sin, and it's the 2nd quarter and in the 2nd quarter cos is negative, so -6sinx-sinx=-7sinx (right?)

Originally Posted by Siron
But this can be simpliefd by usin the fact that $\sin[\arccos(x)]=\sqrt{1-x^2}$
What does [] in this case mean? But I'd have square in this case arccos0.6, no?

8. ## Re: Trigonometrical equation

Is it:
$6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$
or
$-6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$

?

9. ## Re: Trigonometrical equation

Originally Posted by Siron
Is it:
$6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$
or
$-6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$

?
It's $6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)$
Here it is calculated by WolframAlpha
I don't know how to explain but 6sin(pi+x) can be reduced to -6sinx ....

10. ## Re: Trigonometrical equation

Ok, so indeed you get $7\sin[\arccos(0.6)]=7\sqrt{1-(0,6)^2}=-5,6$