Question

In right triangle ABC, with right angle at C, AC=5 and BC=12. Find the value of (sin A)^2+ (sin B)^2

Answer 1

Answer:

(sin A)² + (sin B)² = 1

Step-by-step explanation:

Look at the picture.

$sine=\dfrac{opposite}{hypotenuse}$

We need the hypotenuse. Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

Substitute

$leg=5,\ leg=12,\ hypotenuse=x$

$5^2+12^2=x^2\\\\25+144=x^2\\\\169=x^2\to x=\sqrt{169}\\\\x=13$

For angle A we have

$opposite=12,\ hypotenuse=13$

Substitute:

$\sin A=\dfrac{12}{13}$

For angle B we ahve

$opposite=5,\ hypotenuse=13$

Substitute:

$\sin B=\dfrac{5}{13}$

Substitute to given expression:

$\sin^2A+\sin^2B=\left(\dfrac{12}{13}\right)^2+\left(\dfrac{5}{13}\right)^2=\dfrac{144}{169}+\dfrac{25}{169}=\dfrac{144+25}{169}=\dfrac{169}{169}=1$

You can also see a relationship:

$cosine=\dfrac{adjacent}{hypotenuse}\to \cos A=\dfrac{5}{13}=\sin B$

Therefore

$\sin^2 A+\sin^2B=\sin^2A+\cos^2A$

It's a Pythagorean identity:

$\sin^2x+\cos^2x=1$

Therefore

$\sin^2A+\cos^2A=1\to\sin^2A+\sin^2B=1$