Verify Trigonometric Equation

[nycmath]

See attachment.

[Math Tutor]

Since [tex]C=90^\circ[/tex], the legs are [tex]a[/tex] (opposite [tex]A[/tex]) and [tex]b[/tex] (opposite [tex]B[/tex]), and [tex]c[/tex] is the hypotenuse:

[tex]\sin A=\frac{a}{c},\quad \cos A=\frac{b}{c},\quad \sin B=\frac{b}{c},\quad \cos B=\frac{a}{c}[/tex]

(That is just the fact that A and B are complementary, so [tex]\sin B=\cos A[/tex] and [tex]\cos B=\sin A[/tex].)

Therefore

[tex]\frac{\sin^2 A}{\sin^2 B}=\frac{a^2/c^2}{b^2/c^2}=\frac{a^2}{b^2},\qquad \frac{\cos^2 A}{\cos^2 B}=\frac{b^2/c^2}{a^2/c^2}=\frac{b^2}{a^2}[/tex]

Subtracting and using the common denominator [tex]a^2b^2[/tex]:

[tex]\frac{\sin^2 A}{\sin^2 B}-\frac{\cos^2 A}{\cos^2 B}=\frac{a^2}{b^2}-\frac{b^2}{a^2}=\frac{a^4-b^4}{a^2b^2}[/tex]

which is exactly the required identity. Note that no Pythagorean theorem is needed here, only the definitions of sine and cosine in the right triangle.

[nycmath]

Since [tex]C=90^\circ[/tex], the legs are [tex]a[/tex] (opposite [tex]A[/tex]) and [tex]b[/tex] (opposite [tex]B[/tex]), and [tex]c[/tex] is the hypotenuse:

[tex]\sin A=\frac{a}{c},\quad \cos A=\frac{b}{c},\quad \sin B=\frac{b}{c},\quad \cos B=\frac{a}{c}[/tex]

(That is just the fact that A and B are complementary, so [tex]\sin B=\cos A[/tex] and [tex]\cos B=\sin A[/tex].)

Therefore

[tex]\frac{\sin^2 A}{\sin^2 B}=\frac{a^2/c^2}{b^2/c^2}=\frac{a^2}{b^2},\qquad \frac{\cos^2 A}{\cos^2 B}=\frac{b^2/c^2}{a^2/c^2}=\frac{b^2}{a^2}[/tex]

Subtracting and using the common denominator [tex]a^2b^2[/tex]:

[tex]\frac{\sin^2 A}{\sin^2 B}-\frac{\cos^2 A}{\cos^2 B}=\frac{a^2}{b^2}-\frac{b^2}{a^2}=\frac{a^4-b^4}{a^2b^2}[/tex]

which is exactly the required identity. Note that no Pythagorean theorem is needed here, only the definitions of sine and cosine in the right triangle.

Nicely-done. Perfectly clear.