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If $x < 90^\circ$, such that $\sin x = \dfrac {2}{\sqrt{5}}$ find $\cos (x)$ without using a calculator and hence evaluate $ \dfrac {1+\cos(x)}{1-\sin(x)}$ in the form $m+n\sqrt{5}$, where $m+n\in\Bbb Z^+$

and

If $y < 90^\circ$, such that $\sin y = \dfrac {1}{\sqrt{5}}$ find $\cos (y)$ and $\tan (y)$ without using a calculator and hence evaluate $8 \sin y + \cos y+6\tan y$ in the form $m+n\sqrt{5}$, where m and n are positive integers.

I'm a little lost on how to approach this, I'm so used to these type of questions but at a more difficult level that these two questions confuse me, and I'm 100% that they are not difficult.

So far I've done this but I don't think it's correct:

$$\frac {2}{\sqrt{5}} \frac {\sqrt{5}}{\sqrt{5}} = \frac {2\sqrt{5}}{5}$$

Therefore,

$$x=\sin^{-1}\left(\frac {2\sqrt{5}}{5}\right)+2\pi n$$

$$x=\pi-\sin^{-1}\left(\frac {2\sqrt{5}}{5}\right)+2\pi n$$

Which I don't think they have anything to do with the above...

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    $\begingroup$ Do you know that $\sin^2 x + \cos^2 x =1$ ? $\endgroup$ – Martin R Jul 24 at 19:20
  • $\begingroup$ @MartinR Yes, but I only have $\sin^{1}$ $\endgroup$ – scoff Jul 24 at 19:21
  • $\begingroup$ @scoff, if you have $\sin^1$, how could you find $\sin^2$? $\endgroup$ – wjmccann Jul 24 at 19:22
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    $\begingroup$ But $\sin^2{x}=(\sin{x})^2$.... $\endgroup$ – Eleven-Eleven Jul 24 at 19:22
  • $\begingroup$ You could also use that $\sin{x}=\cos{\left(\frac{\pi}{2}-x\right)}$ and use the difference formula for $\cos{(x-y)}$ $\endgroup$ – Eleven-Eleven Jul 24 at 19:24
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$\sin^2 x + \cos^2 x = 1$ so

$(\frac 2{\sqrt 5})^2 + \cos^2 x=1$ so .....

$\frac 45 + \cos^2 x=1$

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$\cos^2 x = 1-\frac 45 =\frac 15$

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$\cos x = \pm \sqrt {\frac 15} = \pm \frac 1{\sqrt 5}$.

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But $x< 90^\circ$ so $\cos x > 0$.

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So $\cos x = \frac 1{\sqrt 5}$.

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Hint:

We have $$ \cos^2(x) = 1- \sin^2(x) $$ Thus $$ \cos^2(x) = 1 - \frac{4}{5} = \frac{1}{5} $$ Then ($x < 90°$) $$ \cos(x) = \frac{1}{\sqrt{5}} $$ Can you finish?

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  • $\begingroup$ What's left to finish? $\endgroup$ – fleablood Jul 25 at 5:05
  • $\begingroup$ @fleablood "and hence evaluate $\dfrac {1+\cos(x)}{1-\sin(x)}$". $\endgroup$ – Monadologie Jul 25 at 7:19
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$\sin^2{x} + \cos^2{x} = 1$

$\cos^2{x} = 1 - \sin^2{x}$

$\cos{x} = \sqrt{1 - \sin^2{x}} \quad \land \quad \sin{x} = \frac{2}{\sqrt{5}} \quad \land \quad x < 90^{\circ}$

$\Longrightarrow \cos x = \sqrt{1 - \Big(\frac{2}{\sqrt{5}} \Big)^2}=\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}}$

Can you continue?

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  • $\begingroup$ "Can you continue?"? To where? $\endgroup$ – fleablood Jul 24 at 19:37
  • $\begingroup$ To do the other question he means - thank you the three of you $\endgroup$ – scoff Jul 24 at 19:56

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