How is this wrong solution?

[dollyayesha2345]

If [math]sin x=cos x[/math] then find the value of [math]2tanx-cos^2x[/math].

My attempt:
[math]\frac{sin\ x}{cos\ x}=\frac{cos\ x}{cos\ x}[/math][math]tan x =1[/math][math]x=tan^{-1}[/math][math]x=45°[/math]Substituting [math]x=45°[/math] in [math]2tanx-cos^2x[/math][math]2tan45°-cos^2(45°)[/math][math]2tan45°-cos90°[/math] (I received a feedback saying "wrong solution" on this step exactly)
[math]=2(1)-0[/math][math]=2[/math]

 

[BigBeachBanana]

Where's the solution? You should know how the forum works by now. Would you please show your work and tell us why you're stuck?

 

[dollyayesha2345]

Where's the solution? You should know how the forum works by now. Would you please show your work and tell us why you're stuck?

I was typing and accidentally hit "Enter". Now I have edited the post with my attempt.

 

[BigBeachBanana]

2tan45°−cos2(45°)2tan45°−cos90°2tan45°-cos90°2tan45°−cos90°

The error you made is here.
[math]cos^2(45\degree)\neq cos(90\degree)[/math][math]cos^2(45\degree) = cos(45\degree)*cos(45\degree)[/math]

 

[dollyayesha2345]

The error you made is here.
[math]cos^2(45\degree)\neq cos(90\degree)[/math][math]cos^2(45\degree) = cos(45\degree)*cos(45\degree)[/math]

So I guess it should be like this:
[math]2tan45°−cos^2 (45°)[/math][math]=2(1)-\frac{1}{2}[/math][math]=\frac{3}{2}[/math]Right?

 

[BigBeachBanana]

So I guess it should be like this:
[math]2tan45°−cos 2 (45°)[/math][math]=2(1)\times[/math][math]2(1)\times\frac{1}{2}[/math][math]=\frac{3}{2}[/math]Right?

Not quite. Think [imath]cos^2(45\degree)[/imath] similar to [imath]x^2=x*x[/imath]. You evaluate [imath]cos(45\degree)[/imath] first, then square it. That's what the notation means. Another way you can write it as: [math]cos^2(45\degree) = [cos(45\degree)]^2[/math]

 

[dollyayesha2345]

Not quite. Think [imath]cos^2(45\degree)[/imath] similar to [imath]x^2=x*x[/imath]. You evaluate [imath]cos(45\degree)[/imath] first, then square it.

[math]cos^245=(\frac{\sqrt2}{2})^2[/math][math](\frac{\sqrt2}{2})^2=\frac{1}{2}[/math]

 

[BigBeachBanana]

[math]cos^245=(\frac{\sqrt2}{2})^2[/math][math](\frac{\sqrt2}{2})^2=\frac{1}{2}[/math]

Yes, you are right, my apologies. However, you switched the operation from minus to multiply.

 

[dollyayesha2345]

Yes, you are right, my apologies. However, you switched the operation from minus to multiply.

Thanks for your reply!

 

[Deleted member 4993]

x=tan−1(1)

There are at least two solutions → x =\(\displaystyle \frac{\pi}{4} \pm n*\pi \) ........................... n = 0,1,2,3.......................

 

[Steven G]

[math]tan x =1[/math][math]x=tan^{-1}[/math][math]x=45°[/math]

tan^-1 is a function, that is for any input you'll never get back more than one answer.

tan x =1 and x=tan^-1(1) are NOT the same!

tan x = 1 has infinitely many solutions where as x=tan^-1(1) has just one solution.
There is a solution in quadrant I, which you found out to be x=π/4. But that is not complete. x can equal π/4±n∗2π, when n is any integer.
Now is there another quadrant where tan x = 1? What angle would that be?