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

$tan^{2}36\cdot tan^{2}72=$

2. ## Re: Calculate tan^2(36) tan^2(72)

The Problem is very easy.

$\tan^2{18}&space;\tan^{2}72&space;=&space;\tan^{2}18\cot^{2}(90-72)&space;=\tan^{2}18\cot^{2}18=1$

3. ## Re: Calculate tan^2(36) tan^2(72)

no $tan^2(18)tan^2(72)$ I know this is easy

but
$tan^2(36)tan^2(72)$

4. ## 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 $\tan^2{(72)}.\tan^2{(36)}=25-20=5$

5. ## 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

6. ## 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.

yes

8. ## 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.

9. ## Re: Calculate tan^2(36) tan^2(72)

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 ....

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

$\Rightarrow \sin{72}=\frac{\sqrt{10+2\sqrt{5}}}{4}$

11. ## 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).

$\sin{36}=2\sin{18}\cos{18}$

12. ## 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

$Rcos2\theta+iRsin2\theta=(rcos\theta+irsin\theta)^ 2$
$R=rr=r^2$
$R+Rcos2\theta=2rrcos^2\theta$
$R-Rcos2\theta=2rrsin^2\theta$

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### tg^2 36*tg^2 72=5

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