Inverse stuff (1 Viewer)

nrlwinner · Mar 20, 2010

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$2sin^{-1}\frac{3}{5}=sin^{-1}\frac{24}{25}$

and

$sin^{-1}\frac{3}{5}+tan^{-1}\frac{7}{24}=cos^{-1}\frac{3}{5}$

I have no idea where to start in both of them except for drawing triangles.

untouchablecuz · Mar 20, 2010

$\\\sin\left(2\sin^{-1}\frac{3}{5} \right )\\=2\sin \left(\sin^{-1}\frac{3}{5} \right )\cos\left(\sin^{-1}\frac{3}{5} \right )\\=2\sin \left(\sin^{-1}\frac{3}{5} \right )\sqrt{1-\left(\sin\left(\sin^{-1}\frac{3}{5} \right )\right)^2}\\=2\times\frac{3}{5}\times\sqrt{1-\left(\frac{3}{5}\right)^2}\\=\frac{24}{25}\\\therefore 2\sin^{-1}\frac{3}{5}=\sin^{-1} \frac{24}{25}$

i gotta go now, ill do the other one later if its not done

Ferox · Mar 20, 2010

RTP: sin-1(3/5) + tan-1(7/24) = cos-1(3/5)

Let α = sin-1(3/5) and β= tan-1(7/24)

Now, cos(α + β) = cosα.cos β - sinα.sinβ
= 4/5.24/25 – 3/5.7/25
= 3/5

Therefore α + β = cos-1(3/5)
Therefore sin-1(3/5) + tan-1(7/24) = cos-1(3/5) as reqired

adomad · Mar 20, 2010

Ferox said:
RTP: sin-1(3/5) + tan-1(7/24) = cos-1(3/5)

Let α = sin-1(3/5) and β= tan-1(7/24)

Now, cos(α + β) = cosα.cos β - sinα.sinβ
= 4/5.24/25 – 3/5.7/25
= 3/5

Therefore α + β = cos-1(3/5)
Therefore sin-1(3/5) + tan-1(7/24) = cos-1(3/5) as reqired

MS paint FTW:rofl:

Inverse stuff (1 Viewer)

nrlwinner

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untouchablecuz

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Ferox

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adomad

HSC!!

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