Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$

Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.

I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$

So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.

The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.

• So you did all that but you can't find $\tan\left( \dfrac{\pi}{4}\right)$? Jan 1 '14 at 2:01
• Draw an isoceles right-angled triangle. Two of its angles are $\pi/4$. Then $\tan(\pi/4)$ is opposite divided by adjacent. These are equal, so $\tan(\pi/4)=1$. Jan 1 '14 at 2:07
• You don't have to actually compute $\sin^{-1}(\sqrt{2}/2)$ to solve this problem. Draw a right triangle with opposite side length $\sqrt{2}$ and hypotenuse length $2$. Now use the pythagorean theorem to find the length of the adjacent side; call this length $a$. Then compute $\sqrt{2}/a$. Jan 1 '14 at 2:12
• it's amazing that questions like these get more response that the questions which actually deserve attention Jul 10 '19 at 10:52

Hint: $$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}$$

• You mispelled 'answer'. Jan 1 '14 at 2:05
• No, because he didn't also misspell $1$.
– John
Jan 1 '14 at 2:14

$\hskip2in$ Using the triangle above...& the fact that $$\tan x = \frac{\text{opp}}{\text{adj}}, \space \tan \left(\frac{\pi}{4}\right)=...$$

We have to find the exact value of $$\;\tan\left(\sin^{-1}\dfrac{1}{\sqrt{2}}\right)$$.

Since $$\;\sin(45)=\dfrac{1}{\sqrt{2}}\;,\;x=45\;$$ so $$\;\tan(45)=1\;,\;$$ hence we are done. Now $$\;\sin(45)=\cos(45)\;,\;$$ thus $$\;\tan(45)=1\;.$$

You have

$$\sin x=\dfrac{\sqrt{2}}{2}$$

so

$$\cos x=\sqrt{1-(\sin x)^{2}}=\dfrac{\sqrt{2}}{2}$$

and

$$\tan x =\dfrac{\sin x}{\cos x}\ = 1$$