Question: Given: $\triangle ABC$ is a right triangle.
Prove: $(\sin (A))^2 + (\cos (A))^2 = 1$
A
c
B
a
b
C
Drag reasons into the table to complete the proof.
Statement Reason
$\triangle ABC$ is a right triangle Given
$\sin (A) = \frac{a}{c}$
$\cos (A) = \frac{b}{c}$
$a^2 + b^2 = c^2$
$\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$
$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$
$(\sin (A))^2 + (\cos (A))^2 = 1$
Definition of cosine
Definition of sine
Definition of right triangle
Pythagorean Theorem
Exponent rules for division
Substitution Property
Multiplication Property of Equality
Transitive Property of Equality
1. **State the problem:** We need to prove that for a right triangle $\triangle ABC$, the identity $(\sin (A))^2 + (\cos (A))^2 = 1$ holds.
2. **Recall definitions:**
- By definition of sine in a right triangle, $\sin (A) = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{a}{c}$.
- By definition of cosine, $\cos (A) = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{c}$.
3. **Use the Pythagorean theorem:**
Since $\triangle ABC$ is right angled at $C$, the sides satisfy $a^2 + b^2 = c^2$.
4. **Divide the Pythagorean equation by $c^2$:**
$$\frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{c^2}{c^2} = 1$$
5. **Rewrite using exponent rules:**
$$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$$
6. **Substitute sine and cosine definitions:**
$$ (\sin (A))^2 + (\cos (A))^2 = 1 $$
7. **Conclusion:** We have shown step-by-step that $(\sin (A))^2 + (\cos (A))^2 = 1$ using the definitions and the Pythagorean theorem.
This completes the proof.