1. ## θ is Acute?

Can anybody help me with this following question, I never heard of it before...until now:

If tan θ = x where θ is acute, find in the terms of x:

a) 2 sin θ.
b) 3 cos θ.

What does θ acute in this case represent?

2. If $\theta$ is acute then

$\sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta} }$

$\cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}$

3. Originally Posted by red_dog
If $\theta$ is acute then

$\sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta} }$

$\cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}$
Ah thank you for your help!

4. Originally Posted by Lites
Can anybody help me with this following question, I never heard of it before...until now:

If tan θ = x where θ is acute, find in the terms of x:

a) 2 sin θ.
b) 3 cos θ.

What does θ acute in this case represent?
In terms of x...

if $tan\theta=x$

$tan\theta=\frac{x}{1}=\frac{opposite}{adjacent}$ in a right-angled triangle.

Using Pythagoras' theorem

$hypotenuse=\sqrt{adj^2+opp^2}=\sqrt{1+x^2}$

$2sin\theta=2\frac{opp}{hyp}=\frac{2x}{\sqrt{1+x^2} }$

$3cos\theta=3\frac{adj}{hyp}=\frac{3}{\sqrt{1+x^2}}$

5. Originally Posted by Archie Meade
In terms of x...

if $tan\theta=x$

$tan\theta=\frac{x}{1}=\frac{opposite}{adjacent}$ in a right-angled triangle.

Using Pythagoras' theorem

$hypotenuse=\sqrt{adj^2+opp^2}=\sqrt{1+x^2}$

$2sin\theta=2\frac{opp}{hyp}=\frac{2x}{\sqrt{1+x^2} }$

$3cos\theta=3\frac{adj}{hyp}=\frac{3}{\sqrt{1+x^2}}$