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Move expressions and reasons into the blanks to complete the proof of [tex]\sin^2 \theta + \cos^2 \theta = 1[/tex].

According to [tex]$\qquad$[/tex] It is known that [tex]$a^2 + b^2 = 1$[/tex]

By the definition of sine, [tex]$\sin \theta = \qquad$[/tex]

By the definition of cosine, [tex]$\cos \theta = \qquad$[/tex]

Therefore by [tex]$\qquad$[/tex] [tex]$\sin^2 \theta + \cos^2 \theta = 1$[/tex]

- the Pythagorean Identity
- [tex]$a$[/tex]
- [tex]$a$[/tex]
- [tex]$\frac{1}{a}$[/tex]
- [tex]$\frac{1}{a}$[/tex]
- [tex]$\frac{1}{a}$[/tex]
- combining like terms
- substitution
- [tex]$\frac{b}{a}$[/tex]
- [tex]$\frac{a}{b}$[/tex]
- [tex]$b$[/tex]
- [tex]$\frac{1}{b}$[/tex]


Sagot :

Here is the completed proof of [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex] with the correct expressions and reasons filled in:

According to the Pythagorean Identity, it is known that [tex]\(a^2 + b^2 = 1\)[/tex].

By the definition of sine, [tex]\(\sin \theta = \frac{b}{c}\)[/tex].

By the definition of cosine, [tex]\(\cos \theta = \frac{a}{c}\)[/tex].

Therefore, by combining like terms, [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].

So the completed proof is:

According to the Pythagorean Identity it is known that [tex]\(a^2 + b^2 = 1\)[/tex].

By the definition of sine, [tex]\(\sin \theta = \frac{b}{c}\)[/tex].

By the definition of cosine, [tex]\(\cos \theta = \frac{a}{c}\)[/tex].

Therefore, by combining like terms, [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].