Prove Trig Identity: CosθSinθ = Cos2θ+CosθSinθ

AI Thread Summary
The discussion centers on proving the trigonometric identity CosθSinθ = Cos2θ + CosθSinθ. Participants note that the identity fails when substituting θ = 0, as the left side equals 0 while the right side equals 1. This leads to the conclusion that the statement is likely a trick question. Instead of proving the identity, it may be more appropriate to solve the equation, which has solutions at the zeros of cosine, specifically odd integer multiples of π/2. The conversation highlights the importance of verifying identities before attempting to prove them.
Bradyns
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Prove:
\frac{CosθSinθ}{1 + Tanθ} = Cos2θ
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I multiply out the denominator to get:

CosθSinθ = Cos2θ + CosθSinθ

I cannot seem to prove it.

Starting to think it's a trick question.. :/
 
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It most certainly is a trick question because the supposed identity doesn't even work if you plug in theta = 0. The left side is 0 while the right is 1, in this case!
 
You might be asked to solve the equation instead of proving the identity, and the solutions of the equation are the zeros of cosine, id est odd integer multiples of \pi/2.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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