4

Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle.

Show that,

$$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$$

How do I go about proving this identity? Can anybody help? I know of the arctan.

  • if you take the $\sin$ of both sides and use the Pythagoras theorem and trigonometric identities, the answers follows immediately. – Chip May 09 '16 at 04:51
  • How can you take the sin of both sides, is that possible? –  May 09 '16 at 04:53
  • $\sin(\arcsin\frac{b}{c}-\cdots)=\sin2\arcsin(\frac{b-a}{c\sqrt 2})$... – Chip May 09 '16 at 04:56

3 Answers3

3

HINT:

$$\dfrac a{\sin A}=\dfrac b{\sin B}=\dfrac c1$$

$\implies\arcsin\dfrac ac=\arcsin(\sin A)=A$ as $0<A<\dfrac\pi2$

Now $A=\dfrac\pi2-B$

and $\dfrac{b-a}{\sqrt2c}=\dfrac{\sin B-\sin A}{\sqrt2}=\dfrac{\sin B-\cos B}{\sqrt2}=\sin\left(B-\dfrac\pi4\right)$

Can you take it from here?

lab bhattacharjee
  • 274,582
2

Here's a trigonograph (with $a$ and $b$ reversed, just because):

enter image description here

$$\begin{align} 2\;\angle X &\;=\; \stackrel{\frown}{PQ} - \stackrel{\frown}{RS} \\[6pt] \implies \qquad\qquad\qquad 2\;\gamma &\;=\; \alpha - \beta \\[6pt] \implies \qquad 2\,\arcsin \frac{a-b}{2\sqrt{c}} &\;=\; \arcsin\frac{a}{c} - \arcsin\frac{b}{c} \end{align}$$

Blue
  • 75,673
  • That is amazing @Blue. I have another formula like this in term of cosine please show another beautiful construction like the above. It is so beautiful. Thank you for the answer. –  May 09 '16 at 10:03
  • @adambui: Post your cosine formula as a separate question, and I'll see what I can do. :) – Blue May 09 '16 at 10:06
1

  1. $\arcsin \dfrac bc -\arcsin \dfrac ac =2\arcsin \dfrac{b-a}{c \sqrt 2}$

  2. $a,b,c \gt 0$

  3. $a^2 + b^2 = c^2$

Let's rewrite this as $\dfrac 12 \left(\arcsin \dfrac bc -\arcsin \dfrac ac \right) =\arcsin \dfrac{b-a}{c \sqrt 2}$

Since you said that sides $a, b, $ and $c$ are sides of a right triangle, then we know that $a,b,c \gt 0$.

Let $\theta$ be the angle that corresponds to the point $(a,b)$ in the first quadrant.

Then $\sin \theta= \dfrac bc$ and $\cos \theta = \dfrac ac$

So $\dfrac 12 \left( \arcsin \dfrac bc -\arcsin \dfrac ac \right) = \dfrac 12 \left( \theta - \left(\dfrac{\pi}{2} - \theta\right) \right) = \theta - \dfrac{\pi}{4}$

If you can show that $\sin \left( \theta - \dfrac{\pi}{4} \right) = \dfrac{b-a}{c \sqrt 2}$, then you are done.

Steven Alexis Gregory
  • 26,769