trig identity

[renegade05]

Alright, quick question.

How does

\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = (1-\sin(\theta))(1+\sin(\theta))^2\)

I can get as far as applying the trig identity \(\displaystyle \cos^2(\theta) = (1-\sin^2(\theta))\) but can't see where to go from there. :?

 

[Mrspi]

Alright, quick question.

How does

\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = (1-\sin(\theta))(1+\sin(\theta))^2\)

I can get as far as applying the trig identity \(\displaystyle \cos^2(\theta) = (1-\sin^2(\theta))\) but can't see where to go from there. :?

Well....let's work with the right side, maybe (since you already know this, apparently: \(\displaystyle \cos^2(\theta) = (1-\sin^2(\theta))\)

\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = (1-\sin(\theta))(1+\sin(\theta))^2\)
\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = (1-\sin(\theta))(1+\sin(\theta))(1+\sin(\theta))\) Applying the definition of squaring
\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = (1-\sin^2(\theta))[(1+\sin(\theta))\) Multiplying out the sum and product
\(\displaystyle \cos^2(\theta)(1+ \sin(\theta)) = \cos^2(\theta)[(1+\sin(\theta))\)

Ok?

 

[soroban]

Hello, renegade05!

We can start with the left side . . .

\(\displaystyle \text{verify: }\:\cos^2\!\theta(1+ \sin \theta) \:=\1-\sin\theta)(1+\sin\theta)^2\)

\(\displaystyle \text{We have: }\:\underbrace{\cos^2\!\theta}_{\downarrow}(1 + \sin\theta)\)
. . . . \(\displaystyle = \;\underbrace{(1-\sin^2\!\theta)}_{\text{Factor}}(1 + \sin\theta)\)
. \(\displaystyle =\;\overbrace{(1-\sin\theta)(1+\sin\theta)}(1+\sin\theta)\)

. . . . \(\displaystyle =\;(1-\sin\theta)(1+\sin\theta)^2\)

 

[renegade05]

Yes Yes i see now... difference of squares. Of course. So simple !

 

[shadiq]

Thanks for examples in various angle.