Prove that - \sin {}^(2) (\theta) + \cos {}^(2) (\theta) = 1 ty !​

Question

Prove that -

`\sin {}^(2) (\theta) + \cos {}^(2) (\theta) = 1`

ty !​

Answer 1

Explanation

There are many ways to prove this identity. But, I will use the simplest one. If you have a right triangle, the length of the horizontal side is x, the length of the vertical side is y, and the hypotenuse is r. So, the relation between x and r can be written in cosine form:

`\cos\theta=(x)/(r)`

The relation between y and r can be written in sine form:

`\sin\theta=(y)/(r)`

Pythagoras told you that for the right triangle
`x^(2)+y^(2)=r^(2)` should be satisfied. Substitute x and y from the cosine and sine form:

`(r\cos\theta)^(2)+(r\sin\theta)^(2)=r^(2)`

`r^(2)\cos^(2)\theta+r^(2)\sin^(2)\theta =r^(2)`

the square of r will cancel out, and you can get your identity:

`\cos^(2)\theta+\sin^(2)\theta=1`