Proof of inverse trigonometric identities

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To prove that arcsin(1/sqrt(5)) + arcsin(2/sqrt(5)) = Pi/2, consider each term as an angle in a right triangle. For the first term, let theta = arcsin(1/sqrt(5), which gives sin(theta) = 1/sqrt(5). Construct a right triangle with the opposite side equal to 1 and the hypotenuse equal to sqrt(5) to find the adjacent side. Determine if there is an angle in that triangle with a sine of 2/sqrt(5), which would correspond to the second term. The sum of these angles, theta and phi, will equal Pi/2.
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Homework Statement



Show that arcsin(1/sqrt(5)) + arcsine(2/sqrt(5)) = Pi/2

Homework Equations





The Attempt at a Solution



Can someone please give me so much as a hint?
 
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Take the sine of the left-hand side and see if it is equal to sin(pi/2)

ehild
 
seboastien said:

Homework Statement



Show that arcsin(1/sqrt(5)) + arcsine(2/sqrt(5)) = Pi/2

Look at each term individually. Each is the arcsine of a ratio, so each term is an angle.

For the first term, if theta = arcsin(1/√5) , then sin(theta) = 1/√5 . Draw a right triangle with one angle being (theta) and having the side opposite (theta) equal to 1 and the hypotenuse equal to √5 . What is the side adjacent to (theta) equal to?

Now, is there an angle in that triangle having a sine of 2/√5 ? If so, it would be an angle which is the arcsine of (2/√5) . Call it (phi) . What do (theta) and (phi) add up to?
 
seboastien said:

Homework Statement



Show that arcsin(1/sqrt(5)) + arcsine(2/sqrt(5)) = Pi/2

Homework Equations





The Attempt at a Solution



Can someone please give me so much as a hint?

Draw a triangle and use pythagorus. [Just hints, not answers]
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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