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Prove that $\alpha+\beta=\frac{\pi}{2}$, if $\sin{\alpha}=\frac{8}{17}$ and $\sin{\beta}=\frac{15}{17}$, where $0<\alpha<\frac{\pi}{2}$ and $0<\beta<\frac{\pi}{2}$.

This is what I did but I don't like it and thought there might be a "nicer" way of doing it.

Since $\sin{\alpha}=\cos{(\frac{\pi}{2}-\alpha)}$, then $\cos{(\frac{\pi}{2}-\alpha)}=\frac{8}{17}$ and since $\sin{\beta}=\cos{(\frac{\pi}{2}-\beta)}$, then $\cos{(\frac{\pi}{2}-\alpha)}=\frac{15}{17}$.

Solving for $\alpha$ gives us (I used positive $\arccos$ for the sake of simply demonstrating the method):

$$\frac{\pi}{2}-\alpha=\arccos{\frac{8}{17}+2\pi n}$$

$$\frac{\pi}{2}-\beta=\arccos{\frac{15}{17}+2\pi n}$$

adding both equations gives:

$$\alpha+\beta=\pi-\left(\arccos{\frac{8}{17}}+\arccos{\frac{15}{17}}\right)-4\pi n$$ but $$\arccos{\frac{8}{17}}+\arccos{\frac{15}{17}}=\frac{\pi}{2}$$

Therefore,

$$\alpha+\beta=\frac{\pi}{2}$$

Blue
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PomPom
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    Well, if you use the fact that $8,15,17$ are a Pythagorean triplet, you work is nearly done. – AryanSonwatikar Jun 24 '19 at 14:47
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    $17^2=8^2+15^2$ so one can construct a right triangle with lengths representing these sides, this means that... –  Jun 24 '19 at 14:52

5 Answers5

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You have $\sin( {\alpha + \beta}) = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}$, so you could show that $\sin (\alpha + \beta) = 1$ which would lead to $\alpha + \beta = \frac{\pi}{2}$, if you know the values for $\cos \alpha$ and $\cos \beta$.

Aden
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Hint for nicer way: Do the given sine values allow $\alpha$ and $\beta$ to be the two non-right angles of a right-angled triangle? (Let the hypothetical hypothenuse be $17$ for simplicity.)

Arthur
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  • You mean, I can find out that $\cos{\alpha}=\frac{15}{17}$ and $\cos{\beta}=\frac{8}{17}$? This is how I started but did not know where to go from there really. – PomPom Jun 24 '19 at 14:57
  • @PomPom Forget cosine. If there is a right-angled triangle with hypotenuse $17$, and angles $\alpha$ and $\beta$, what does the sine values tell you that the lengths of the legs must be? Do those two legs and that hypotenuse together really make a right-angled triangle? – Arthur Jun 24 '19 at 15:27
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You’ve shown that $\sin\alpha=\cos\beta=\cos\left(\frac\pi2-\alpha\right)$ and $\cos\alpha = \sin\beta = \sin\left(\frac\pi2-\alpha\right)$. Since both of these angles are in the first quadrant, you can conclude directly that $\beta = \frac\pi2-\alpha$. Otherwise, you could use the angle addition formulas: $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\cos\beta = \sin^2\alpha+\cos^2\alpha=1 \\ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos\alpha\sin\alpha - \sin\alpha\cos\alpha = 0.$$

amd
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You could calculate $\cos(\alpha)$ and $\cos(\beta)$ from knowing they’re in the first quadrant and using $\sin^2+\cos^2=1$. Then use the angle addition formula to calculate $\sin(\alpha+\beta)$. What can you conclude?

J. W. Tanner
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Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

Here $x^2+y^2=1$

Alternatively

If $\arcsin x=A$

$\cos A=+\sqrt{1-x^2}$

$\tan A=?,A=\arctan(?)$

Finally use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

See also Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

lab bhattacharjee
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