Another Trig Identity

by cscott
Tags: identity, trig
 P: 786 $$1 - \frac{\cos^2 \theta}{1 + \sin^2 \theta} = \sin \theta$$ I can only get to $$\frac{2\sin^2 \theta}{1 + \sin^2 \theta}$$ and I don't know if that's correct.
 HW Helper P: 1,024 Perhaps you should check the problem again because it doesn't seem correct to me. Try $\theta = 3\pi /2$, the LHS will be 1 while the RHS is -1.
 HW Helper P: 878 It's not an identity! Let θ = -π/2 then lhs = 1 - cos²(-π/2)/(1 + sin²(-π/2)) lhs = 1 - 0/2 lhs = 1 ===== but rhs = sin(-π/2) = -sin(π/2) = -1 rhs = - 1 ======= Since lhs ≠ rhs, it can't be an identity.
 P: 786 Another Trig Identity OMG, I've spent so much time trying to figure it out... I never thought of checking to make sure it was indeed an identity. I guess my teacher made a mistake. Thanks guys. (Don't worry, I'll be back with another too soon )
 HW Helper P: 878 I did a quick check to see if it was an identity. After all , you were having trouble ! It checked out ok for θ = 0 and θ = pi/2, so I did some work on it. It was only later on that I plugged the expressions into graphmatica and found out they were actually two different functions !!
 P: 786 $$\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta$$ $$\frac{1 - \sin \theta}{1 + \sin \theta} = (\sec \theta - \tan \theta)^2$$ I can't seem to get anywhere with these...
 HW Helper P: 878 For the 1st one, cross-multiply the two terms in the lhs and simplify.
 HW Helper P: 878 For the 2nd one, multiply the lhs by (1 - sinθ) and simplify.
 P: 786 I can only get to $2 \sin \theta + 2 \sin^2 \theta$ for the first after cross multiplying. For the second, how can I just multiply by $(1 - \sin \theta)$? I thought you can only multiply by 1?
HW Helper
PF Gold
P: 12,016
 Quote by cscott $$\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta$$ $$\frac{1 - \sin \theta}{1 + \sin \theta} = (\sec \theta - \tan \theta)^2$$ I can't seem to get anywhere with these...
$$\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta}=\frac{1 - \sin \theta}{\cos \theta}+\frac{1+\sin\theta}{1+\sin\theta}\frac{\cos\theta}{1-\sin\theta}=\frac{1 - \sin \theta}{\cos \theta} +\frac{1+\sin\theta}{\cos\theta}=\frac{2}{\cos\theta}=2sec\theta$$

$$\frac{1 - \sin \theta}{1 + \sin \theta} =\frac{1-\sin\theta}{1-\sin\theta}\frac{1-\sin\theta}{1+\sin\theta}=\frac{(1-\sin\theta)^{2}}{\cos^{2}\theta}=(sec\theta-tan\theta)^{2}$$
 P: 786 I see, thanks.
HW Helper
P: 878
 Quote by cscott I can only get to $2 \sin \theta + 2 \sin^2 \theta$ for the first after cross multiplying. For the second, how can I just multiply by $(1 - \sin \theta)$? I thought you can only multiply by 1?
Ah, sorry, that should have been, multiply above and below by (1 - sinθ)

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