Questions about expressions of trigonometry

Questions
,
1

I’ve just started to explore Wolfram Alpha and I’m very fascinated. I’ve a simple a question for you:
expressions like this

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

2

Hi,

yes, for the step by step solution you need a pro subscription.

You might want to try a free Wolfram Programming Cloud account

http://www.wolfram.com/programming-cloud/pricing/

which comes with some free Wolfram API requests and should show step by step solutions.

Cheers,

M.

3

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

4

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].

5

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