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Maths Trig Identities Help

Hi - can anyone explain how you do this to me? I have to root sin^2theta/sin theta divided by cos theta
But unsure where to go from there: It is Q8 on the attachment

Thanks for your help!
(edited 8 years ago)
maths2.png106.0KB
it is not entirely clear what your expression looks like... but it may cancel down to

√ ( tan θ )
uh could you write that out a little clearer please? Like, write down the problem using these:

sin^2 x

cos x
I think you cancel the sin^2theta/sin theta down to just sin theta, and then that divided by cos theta will give you tan theta but not 100% sure sorry!
Reply 4
Avatar for Zacken
Zacken
22
Original post by Shadowshock118
Hi - can anyone explain how you do this to me? I have to root sin^2theta/sin theta divided by cos theta
But unsure where to go from there: It is Q8 on the attachment

Thanks for your help!


So, right off the bat: 1−cos⁡2θ≡sin⁡2θ1 - \cos^2 \theta \equiv \sin^2 \theta so that 1−cos⁡2θ=sin⁡2θ=sin⁡θ\sqrt{1-\cos^2 \theta} = \sqrt{\sin^2 \theta} = \sin \theta.

This means you have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\sqrt{1-\cos^2 \theta}}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \times \frac{1}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \times \frac{\cos \theta}{\sin \theta} = \cdots\end{equation*}



Moved to maths, by the way.
(edited 8 years ago)
Original post by sammyyrosee
uh could you write that out a little clearer please? Like, write down the problem using these:

sin^2 x

cos x


I got to root sin^2 x/ sinx ÷ cosx
Original post by Zacken
So, right off the bat: 1−cos⁡2θ≡sin⁡2θ1 - \cos^2 \theta \equiv \sin^2 \theta so that 1−cos⁡2θ=sin⁡2θ=sin⁡θ\sqrt{1-\cos^2 \theta} = \sqrt{\sin^2 \theta} = \sin \theta.

This means you have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\sqrt{1-\cos^2 \theta}}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \times \frac{1}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \times \frac{\cos \theta}{\sin \theta} = \cdots\end{equation*}



Moved to maths, by the way.


Brilliant, thank you very much
Reply 7
Avatar for Zacken
Zacken
22
Original post by Shadowshock118
I got to root sin^2 x/ sinx ÷ cosx


Yes, and sin⁡2x=(sin⁡2x)1/2=(sin⁡x)1/2×2=(sin⁡x)1=sin⁡x\displaystyle \sqrt{\sin^2 x} = (\sin^2 x)^{1/2} = (\sin x)^{1/2 \times 2} = (\sin x)^1 = \sin x .

Edit: because x is acute, obviously.
(edited 8 years ago)

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