In triangle ABC , if angle c=90°,then prove that sin^2 A+sin^2 B=1.
This basic proof should be found in your textbook or in your notes.
If not, just Google it, you will found many proofs.
It's just the Pythagorean Theorem in disguise.
To prove sin^2 A + sin^2 B = 1 in a right-angled triangle ABC, where angle C = 90°, let's consider the definitions and properties of the trigonometric functions.
We know that the sum of the angles of any triangle is 180°. In triangle ABC, angle C is 90°, which means angle A + angle B = 180° - 90° = 90°.
Now, let's define the trigonometric functions in terms of the sides of a right-angled triangle:
1. Sine (sin) of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse.
sin A = BC/AC
sin B = AC/BC
2. Cosine (cos) of an angle is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.
cos A = AB/AC
cos B = AB/BC
3. Pythagorean Theorem: In a right-angled triangle, the square of the length of the hypotenuse (the longest side) equals the sum of the squares of the lengths of the other two sides.
AC^2 = AB^2 + BC^2
Now, let's start the proof:
1. From the definition of sine, we have sin A = BC/AC and sin B = AC/BC.
2. Square both equations: (sin A)^2 = (BC/AC)^2 and (sin B)^2 = (AC/BC)^2.
3. From the Pythagorean theorem, we have AC^2 = AB^2 + BC^2. Rearranging it, we get BC^2 = AC^2 - AB^2.
4. Plugging the value of BC^2 into the square of sin A equation:
(sin A)^2 = (AC^2 - AB^2)/AC^2.
5. Simplifying the equation: (sin A)^2 = 1 - (AB^2/AC^2).
6. Similarly, plugging the value of AC^2 into the square of sin B equation:
(sin B)^2 = (AB^2 + BC^2)/BC^2.
7. Simplifying the equation: (sin B)^2 = (AB^2 + AC^2)/BC^2.
8. Since AC^2 = AB^2 + BC^2 (from the Pythagorean theorem), we can substitute this value in both equations.
9. We get: (sin A)^2 = 1 - (AB^2/AC^2) and (sin B)^2 = 1 - (AB^2/AC^2).
10. Adding both equations together: (sin A)^2 + (sin B)^2 = 1 - (AB^2/AC^2) + 1 - (AB^2/AC^2) = 2 - 2(AB^2/AC^2).
11. Since AB^2/AC^2 represents a ratio of the lengths of two sides in the right-angled triangle, AB^2/AC^2 = (AB/AC)^2 = (cos A)^2.
12. Substituting this value: (sin A)^2 + (sin B)^2 = 2 - 2(cos A)^2.
13. From the trigonometric identity sin^2 A + cos^2 A = 1, we have sin^2 A = 1 - cos^2 A.
14. Thus, we can rewrite the equation: (sin A)^2 + (sin B)^2 = 2 - 2(cos A)^2 = 2 - 2(1 - (sin A)^2) = 2 - 2 + 2(sin A)^2 = 2(sin A)^2.
15. Therefore, (sin A)^2 + (sin B)^2 = 2(sin A)^2 = 1, which proves the equation sin^2 A + sin^2 B = 1 in a right-angled triangle where angle C = 90°.