# Calculate tan^2(36) tan^2(72)

• Sep 24th 2011, 02:15 AM
afrim1
Calculate tan^2(36) tan^2(72)
$tan^{2}36\cdot tan^{2}72=$
• Sep 24th 2011, 06:46 AM
sbhatnagar
Re: Calculate tan^2(36) tan^2(72)
• Sep 24th 2011, 07:31 AM
afrim1
Re: Calculate tan^2(36) tan^2(72)
no $tan^2(18)tan^2(72)$ I know this is easy

but
$tan^2(36)tan^2(72)$
• Sep 24th 2011, 07:46 AM
sbhatnagar
Re: Calculate tan^2(36) tan^2(72)
Okay sorry I read $\tan^2(18)$

$\tan^2{(36)}=5-2 \sqrt{5}$

also, $\tan^2{(72)}=5+2 \sqrt{5}$

Therefore $\tan^2{(72)}.\tan^2{(36)}=25-20=5$
• Sep 24th 2011, 08:19 AM
afrim1
Re: Calculate tan^2(36) tan^2(72)
hahah very quickly I dont understad where you get worth of tan^2(36) tan^(72) can you explain
• Sep 24th 2011, 10:30 AM
HallsofIvy
Re: Calculate tan^2(36) tan^2(72)
He multiplied! $(tan^2(36))(tan^2(72))= (5- 2\sqrt{5})(5+ 2\sqrt{5})= 25- 4(5)= 5$.

Now, if you are asking how sbhatnagar determined that $tan^2(36)= 5- 2\sqrt{5}$, that's a really good question! Come on, sbhatnagar, let us in on the secret.
• Sep 25th 2011, 01:34 AM
afrim1
Re: Calculate tan^2(36) tan^2(72)
yes
• Sep 25th 2011, 04:27 AM
piscoau
Re: Calculate tan^2(36) tan^2(72)
triple angle formula can be used to find the exact value of tan 36 and tan 72.(Happy)
• Sep 25th 2011, 04:43 PM
mr fantastic
Re: Calculate tan^2(36) tan^2(72)
Quote:

Originally Posted by HallsofIvy
He multiplied! $(tan^2(36))(tan^2(72))= (5- 2\sqrt{5})(5+ 2\sqrt{5})= 25- 4(5)= 5$.

Now, if you are asking how sbhatnagar determined that $tan^2(36)= 5- 2\sqrt{5}$, that's a really good question! Come on, sbhatnagar, let us in on the secret.

To calculate tan(36 degrees), I'd consider a regular pentagon ....
• Oct 6th 2011, 01:42 AM
sbhatnagar
Re: Calculate tan^2(36) tan^2(72)
To find sin(72):

Let x = 18

$5x=90$

$\Rightarrow 3x+2x=90$

$\Rightarrow 2x=90-3x$

$\Rightarrow \sin{2x}=\sin{(90-3x)}$

$\Rightarrow 2\sin{x}\cos{x}=\cos{3x}$

$\Rightarrow 2\sin{x}\cos{x}=4\cos^3{x}-3\cos{x}$

$\Rightarrow 4\sin^2{x}+2\sin{x}-1=0$

So now using quadratic formula, $\Rightarrow \sin{x}=\frac{\sqrt{5}-1}{4}$

Now use sin^(2)x+cos^(2)x=1 to find cos(x)

$\cos{x}=\frac{\sqrt{10+2\sqrt{5}}}{4}$

Now you have, $\cos{18}=\frac{\sqrt{10+2\sqrt{5}}}{4}$

Therefore $\sin{(90-18)}=\frac{\sqrt{10+2\sqrt{5}}}{4}$

$\Rightarrow \sin{72}=\frac{\sqrt{10+2\sqrt{5}}}{4}$
• Oct 11th 2011, 04:26 AM
sbhatnagar
Re: Calculate tan^2(36) tan^2(72)
If $\sin{72}=\frac{\sqrt{10+2\sqrt{5}}}{4}$ then $\cos{72}=\frac{\sqrt{5}-1}{4}$ and $\tan{72}=\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}}$

Therefore $\tan^2{72}=5+2\sqrt{5}$

The value of sin(36 degrees) can be found by using the formula of sin(2x).

$\sin{36}=2\sin{18}\cos{18}$
• Oct 11th 2011, 05:57 AM
BookEnquiry
Re: Calculate tan^2(36) tan^2(72)
ha ha I searched a lot of similar crazy expressions. It seems difficult to learn all of them.
Exact trigonometric constants - Wikipedia, the free encyclopedia
Trigonometry Angles -- from Wolfram MathWorld
• Oct 11th 2011, 06:06 AM
BookEnquiry
$Rcos2\theta+iRsin2\theta=(rcos\theta+irsin\theta)^ 2$
$R=rr=r^2$
$R+Rcos2\theta=2rrcos^2\theta$
$R-Rcos2\theta=2rrsin^2\theta$