Message Boards Message Boards

0
|
10054 Views
|
4 Replies
|
3 Total Likes
View groups...
Share
Share this post:

Questions about expressions of trigonometry

Posted 9 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

POSTED BY: Pietro Rampini
4 Replies

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 BY: Joshua Champion

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 BY: David Gathercole

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 BY: Joshua Champion

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.

POSTED BY: Marco Thiel
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract