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# Questions about expressions of trigonometry

Posted 10 years ago
 I've just started to explore Wolfram Alpha and I'm very fascinated. I've a simple a question for you: expressions like this http://wolfr.am/4msK~0Gn tan^2(7pi/2+a)[1-(sin(a-pi/2))^2]+(cos(3pi/2+a))^2  gaves 1, but how can I view the step by step solution? Does I need Pro? If I download your app for iOS are the Pro features included? Thanks in advance Pietro Rampini
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Posted 10 years ago
 I must ask, did this help you any at all? I can see that some things would reduce, but I do not see how it will always become one with the letter a in there. I don't see what the function of the letter a is for, but I can only assume that it is a variable with its own specific function [much like the letter h and k for shifts along the x and y axis respectively].
Posted 10 years ago
 This looks like homework, possibly written to encourage interpretation of phase shifts in trigonometric expressions.tan^2(7pi/2+a)[1-(sin(a-pi/2))^2]+(cos(3pi/2+a))^2 TeXses out like so: $$\tan ^2\left(\frac{7 \pi }{2}+a\right) \left(1-\sin ^2\left(a-\frac{\pi }{2}\right)\right)+\cos ^2\left(\frac{3 \pi }{2}+a\right)$$ Simplifying via: $\tan(x+\pi / 2)=\tan(x)^{-1}$, $\cos^2(x)+sin^2(x)=1$, $\sin(x+\pi/2)=\cos(x)$. $\tan ^2\left(\frac{7 \pi }{2}+a\right)=\tan^{-2}(a)$ $1-\sin ^2\left(a-\frac{\pi }{2}\right)=1-\cos ^2(a)=\sin^2(a)$ $\cos ^2\left(\frac{3 \pi }{2}+a\right)=\sin ^2(a)$ $$\tan^{-2}(a)\sin^2(a)+\sin ^2(a)=\cos^{2}(a)+\sin ^2(a)=1$$
Posted 10 years ago
 Well, I don't entirely know how to describe this, however I do know that 1-Sin^2(x) is equal to Cos^2(x), and that Tan(7Pi/2) is equal to Tan(Pi/2) so that kinda helps that by reducing to simpler terms Because Cos(a-Pi/2) is also equal to Cos(Pi/2-a). Also, Cos(3Pi/2+a) is equal to -Cos(Pi/2+a). Tan(Pi/2+a) is equal to Sin(Pi/2+a)/Cos(Pi/2+a). I just hope this helps you out, sorry if this doesn't help at all Xd
Posted 10 years ago
 Hi,yes, for the step by step solution you need a pro subscription.You might want to try a free Wolfram Programming Cloud accounthttp://www.wolfram.com/programming-cloud/pricing/which comes with some free Wolfram API requests and should show step by step solutions. Cheers,M.
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