This is a tricky trig problem I'm stuck with. The problem is asking me to simplify $$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$$ to $\sin x\cos x$.

What I've been doing so far is trying to remove those $tan$ functions.
$$1-\left(\frac{\sin^2x\tan x-\cos^2}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\frac{\sin x}{\cos x}+1}\right)$$ I made this complicated. Is there a simple way to do this problem?

  • $\begingroup$ MathJax hint: if you put a backslash before common functions you get the right font and spacing, so \sin x gives $\sin x$ instead of sin x which gives $sin x$ $\endgroup$ – Ross Millikan May 8 '18 at 14:04
  • $\begingroup$ Noted! Still learning how to use the MathJax syntax. $\endgroup$ – AugieJavax98 May 8 '18 at 14:06

We have

$$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1} =1-\frac{\sin^2x\tan x+\cos^2 x}{\tan x+1}=1-\frac{\sin^3x+\cos^3 x}{\sin x + \cos x}=\\=\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2 x)}{\sin x + \cos x}=\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\sin x \cos x$$

  • $\begingroup$ I am confused with the last part of your working. Did you use another identity? $\endgroup$ – AugieJavax98 May 8 '18 at 14:50
  • $\begingroup$ Sorry I skip a step, that is $$\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\frac{\sin x \cos x(\cos x+\sin x)}{\sin x + \cos x}=\sin x \cos x$$ $\endgroup$ – user May 8 '18 at 14:51
  • $\begingroup$ Oh! I see! What tips can you give for tackling such trig problems? $\endgroup$ – AugieJavax98 May 8 '18 at 14:55
  • $\begingroup$ Try to learn the way to obtain and proof all the main trigonometric identities and do a lot of exercises on that! Do not try to memorize all the expressions but try to understand them and check them by simple cases. That's always a good reference en.wikipedia.org/wiki/List_of_trigonometric_identities Bye! $\endgroup$ – user May 8 '18 at 15:00
  • 1
    $\begingroup$ Noted! I guess practice makes perfect. Thanks again $\endgroup$ – AugieJavax98 May 8 '18 at 15:02

You dropped a sign in the first line. The $\cos^2 x$ term should be positive inside the parentheses. Now multiply the numerator and denominator by $\cos x$ and you have a sum of cubes in the numerator. The denominator cancels.

  • $\begingroup$ The signs always get the best of me. $\endgroup$ – AugieJavax98 May 8 '18 at 14:55
  • $\begingroup$ Thanks for the hint $\endgroup$ – AugieJavax98 May 8 '18 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.