# If $x < 90^\circ$, such that $\sin x = \frac {2}{\sqrt{5}}$, find $\cos (x)$

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...

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

$$\sin^2 x + \cos^2 x = 1$$ 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 $$\cos x = \frac 1{\sqrt 5}$$.

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?

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

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

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

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

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

Can you continue?

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