1. Trigonometry exam

So there is this problem
If sinα*cosα = 1/2,
and α ranges from (π ; 3π/2),
how much is sinα + cosα?
Thanks a lot!

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2. Re: Trigonometry exam

$\sin(a)\cos(a) = \dfrac 1 2$

apply magic identity $\sin(2x)=2\sin(x)\cos(x)$

$\dfrac 1 2 \sin(2a) = \dfrac 1 2$

$\sin(2a) = 1$

$2a = \left\{\dfrac \pi 2 ,\dfrac{5\pi}{2}\right\}$

$a = \left\{\dfrac \pi 4, \dfrac{5\pi}{4}\right\}$

3. Re: Trigonometry exam

Originally Posted by romsek
$\sin(a)\cos(a) = \dfrac 1 2$

apply magic identity $\sin(2x)=2\sin(x)\cos(x)$

$\dfrac 1 2 \sin(2a) = \dfrac 1 2$

$\sin(2a) = 1$

$2a = \left\{\dfrac \pi 2 ,\dfrac{5\pi}{2}\right\}$

$a = \left\{\dfrac \pi 4, \dfrac{5\pi}{4}\right\}$
Thanks!

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4. Re: Trigonometry exam

Originally Posted by carrie9987
Thanks!

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I do hope that you weren't taking the exam at the time... :/

5. Re: Trigonometry exam

What I would do: $\displaystyle (sin(a)+ cos(a))^2= sin^2(a)+ 2sin(a)cos(a)+ cos^2(a)= 1+ 2sin(a)cos(a)$. Since we are given that $\displaystyle sin(a)cos9(a)= \frac{1}{2}$, $\displaystyle (sin(a)+ cos(a))^2= 1+ 1= 2$ so $\displaystyle sin(a)+ cos(a)$ is either $\displaystyle \sqrt{2}$, or $\displaystyle -\sqrt{2}$.

Between $\displaystyle \pi$ and $\displaystyle 3\pi/2$, both sin(x) and cos(x) are negative so
$\displaystyle sin(x)+ cos(x)= -\sqrt{2}$.

6. Re: Trigonometry exam

Originally Posted by HallsofIvy
What I would do: $\displaystyle (sin(a)+ cos(a))^2= sin^2(a)+ 2sin(a)cos(a)+ cos^2(a)= 1+ 2sin(a)cos(a)$. Since we are given that $\displaystyle sin(a)cos9(a)= \frac{1}{2}$, $\displaystyle (sin(a)+ cos(a))^2= 1+ 1= 2$ so $\displaystyle sin(a)+ cos(a)$ is either $\displaystyle \sqrt{2}$, or $\displaystyle -\sqrt{2}$.

Between $\displaystyle \pi$ and $\displaystyle 3\pi/2$, both sin(x) and cos(x) are negative so
$\displaystyle sin(x)+ cos(x)= -\sqrt{2}$.
Thank you!

7. Re: Trigonometry exam

Originally Posted by romsek
I do hope that you weren't taking the exam at the time... :/
Haha, i wasn't, it's in two days